Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is the algebraic intuition behind Vieta jumping in IMO1988 Problem 6? Problem 6 of the 1988 International Mathematical Olympiad notoriously asked:
Let $a$ and $b$ be positive integers and $k=\frac{a^2+b^2}{1+ab}$. Show that if $k$ is an integer then $k$ is a perfect square.
The usual way to show this involves a technique called Vieta jumping. See Wikipedia or this MSE post.
I can follow the Vieta jumping proof, but it seems a bit strained to me. You play around with equations that magically work out at the end. I don't see how anyone could have come up with that problem using that proof.
Is there a natural or canonical way to see the answer to the problem, maybe using (abstract) algebra or more powerful tools? In addition, how can someone come up with a problem like this?
| I don't have a strong math background, I think my solution is more suitable for the Math olympiad contest as a young student.
Problem Definition:
Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$. Show that
$\frac{a^2+b^2}{ab+1}$
is the square of an integer.
Proof:
Let $ t = \frac{a^2+b^2}{ab+1}$.
W.L.O.G we assume $1 \le a \le b$.
If $t=1$, we are done.
If $a=b$, we have $2 b^2 = t (b^2+1) \ge 2 (b^2+1)$, contradiction.
If $a = 1$, we have $1+b^2 = t(b+1)$.
As $t \equiv 1 (mod~b)$, we have $t \ge 1+b$, contradiction.
Therefore, the major focus is on $t>1$ and $1 < a < b$.
Since $a^2 \equiv t ~(mod~b)$, we define $k$ such that $a^2 = t + k b$, hence $k$ is an integer and $k < a$.
Note that
$ t = \frac{a^2+b^2}{ab+1} < \frac{a}{b} + \frac{b}{a} \le 1 + b/2 \le b $.
If $k<0$, we have $a^2 = t+kb \le t - b \le 0$, contradiction.
If $k=0$, we have $t = a^2$, done.
Hence, the only remaining case is $0<k<a$.
In this case, we claim that
$$ t = \frac{k^2+a^2}{k a+1} $$
Substitute $t$, we have $ a^2+b^2 = (a^2 - k b)(ab+1)$, hence
$$ b = (a^2 - k b)a - k = a t - k$$
As a result, $k+b = at$ and $kb = a^2-t$, that is, $k$ and $b$ are the solution of the equation
$$ x^2 - at x + (a^2-t) = 0 $$
One solution is $x = b$, the other solution is $x = k$, therefore the claim follows.
Consequently, whenever a feasible pair $(a,b)$ exists, we create another feasible pair $(k,a)$.
By replacing $(a,b)$ by $(k,a)$ the pair become smaller as $0<k<a<b$.
Because all the other case will result in $t$ is a square number, hence by repeating such replacement, we will finally get into the case where $t$ is a square number.
\qed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "104",
"answer_count": 6,
"answer_id": 3
} |
Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it with Mathematica, but I can't figure out how to prove it.
The integral seems quite problematic to me. If the limits were finite, I would do this:
$$\frac{1}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{1}{a^4-b^4}(\sqrt{x^4+a^4}-\sqrt{x^4+b^4})$$
Then, for one of the integrals we will have:
$$\int_A^B \sqrt{x^4+a^4} dx=a^3 \int_{A/a}^{B/a} \sqrt{1+t^4} dt$$
This integral is complicated, but quite well known.
On the other hand $\int_0^\infty \sqrt{1+t^4}dt$ diverges, so I can't consider the two terms separately.
But the integral behaves like I can! If we look at the final expression, it seems like $\int_0^\infty \sqrt{1+t^4}dt=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}}$ even though it can't be correct.
I have to somehow arrive at Beta function, since we have a squared Gamma as an answer.
I'm interested in this integral, since it represents another kind of mean for two numbers $a$ and $b$. If we scale it appropriately:
$$I(a,b)=\frac{8 \sqrt{\pi}}{\Gamma(1/4)^2 } \int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}= \frac{4}{3} \frac{a^2+ab+b^2}{a^3+ab(a+b)+b^3}$$
So, $1/I(a,b)$ is a mean for the two numbers.
| Another way to split the integrals is to replace the square root by a power $p$:
$$\sqrt{x^4 + a^4}\longrightarrow \left(x^4 + a^4\right)^{p}$$
The integral of the separate terms will then converge for $p<-\frac{1}{4}$ and can be expressed in terms of the beta-function. You can substitute $p = \frac{1}{2}$ in the final answer, despite the individual integrals not converging by invoking analytic continuation.
To get to the beta-functions, you can substitute $x = a t$ to get $a$ out of the way, then $u = t^4 + 1$ and finally $u = \frac{1}{v}$ will yield the explicit beta-function form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.
I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?
There is a similar less complicated pattern for computing $x^2$ values.
| To test your hypothesis you could work out the form of the differences from the first few cases.
\begin{align*}
1^{3}-0^{3}&=1\\
2^{3}-1^{3}&=7\\
3^{3}-2^{3}&=19\\
4^{3}-3^{3}&=37
\end{align*}
For example rewrite out $(37-19)-(19-7)=18-6=6$ as:
\begin{align*}
\{(4^{3}-3^{3})-(3^{3}-2^{3})\}&-\{(3^{3}-2^{3})-(2^{3}-1^{3})\}\\
&=(4^{3}-2\cdot3^{3}+2^3)-(3^{3}-2\cdot2^{3}+1^{3})\\
&=4^{3}-3\cdot3^{3}+3\cdot2^3-1^{3}\qquad (\star)\\
&=6
\end{align*}
So you have to find the difference of two differences to get to $6$ (this is called a finite difference pattern, and you have to iterate twice to get the result of $6$ for all such differences, any further iteration ending in a $0$). Now check that pattern $(\star)$ holds in general for some integer $k\ge3$:
\begin{align*}
k^{3}&-3\cdot(k-1)^{3}+3\cdot(k-2)^3-(k-3)^{3}\\
&=k^{3}-3(k^2-3k^2+3k-1)
+3(k^3-2\cdot3k^2+2^2\cdot3k-2^3)
-(k^3-3\cdot3k^2+3^2\cdot3k-3^3)\\
&=\ \ k^3\\
&\ -3k^3\ +\ 9k^2\ -\ 9k\ +\ 3\\
&\ +3k^3-18k^2+36k-24\\
&\ \ -k^3\ +\ \ 9k^2-27k+27\\
&=6
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 2
} |
BMO2 1995 Question 1 - Solve the equation for positive integers (a,b,c) Find all triples of positive integers (a, b, c) such that
$$ \left( 1 + \frac{1}{a} \right)\left( 1 + \frac{1}{b} \right)\left( 1 + \frac{1}{c} \right) = 2 $$
I tried to expand, yielding $ a+b+c+ab+bc+ca+1 = abc $, but I couldn't go anywhere from here.
This problem clearly involves using the fact that $a, b, c$ are positive integers, since there are three variables and only one equation. It would be great if someone also addressed this condition and maybe tried to solve the problem for other requirements, like when $a, b, c$ are positive or negative integers.
Finally - can anyone post or link other such problems which involves a mix of algebra manipulation and number theory. I would really appreciate it.
(Here is the full paper: https://bmos.ukmt.org.uk/home/bmo2-1995.pdf)
| Considering cases works pretty well. If the product is to be $2$, then at least one of the numbers must be less than $4$, and all numbers are clearly $>1$.
Order the numbers so that $a\le b\le c$. Then $a = 2$ or $a=3$. If $a=2$, then $4\le b\le 6$ (since $b=2$ or $b=3$ produces a product too large, and if $b=7$ then $c\ge 7$ and the product is too small). This gives the three solutions $(2,4,15)$, $(2,5,9)$, and $(2,6,7)$. Alternatively, if $a=3$, a similar analysis yields the solutions $(3,3,8)$ and $(3,4,5)$.
So there are five essentially different solutions: $(2,4,15)$, $(2,5,9)$, $(2,6,7)$, $(3,3,8)$, and $(3,4,5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrals $\int\limits _{0}^{\pi/4}\frac{dx}{1+\cos^2x},~~\int\limits_{0}^{\pi/4}\frac{1-\cos^2x}{1+\cos^2x}dx$
Evaluate the definite integrals:
$$I_1=\int\limits _{0}^{\pi/4}\frac{dx}{1+\cos^2x},~~I_2=\int\limits_{0}^{\pi/4}\frac{1-\cos^2x}{1+\cos^2x}dx.$$
Thanks for the help.
Attempt. The classic substitution $\tan(\frac{x}{2})=t$ yields
$$dx=\frac{2dt}{1+t^2},~\cos x=\frac{1-t^2}{1+t^2}$$ and seems to be making the integral more complicated.
| HINT:
*
*Substitute $u=\tan(x)$ and $\text{d}u=\sec^2(x)\space\text{d}x$:
$$\int\frac{1}{1+\cos^2(x)}\space\text{d}x=\int\frac{\sec^2(x)}{1+\sec^2(x)}\space\text{d}x=\int\frac{\sec^2(x)}{2+\tan^2(x)}\space\text{d}x=\int\frac{1}{2+u^2}\space\text{d}u$$
*Substitute $s=\cot(x)$ and $\text{d}s=-\csc^2(x)\space\text{d}x$:
$$\int\frac{1-\cos^2(x)}{1+\cos^2(x)}\space\text{d}x=\int\frac{1-\cos^2(x)}{1+\cos^2(x)}\cdot\frac{-\csc^4(x)}{-\csc^4(x)}\space\text{d}x=$$
$$\int\frac{\csc^2(x)}{1+3\cot^2(x)+2\cot^4(x)}\space\text{d}x=\int\frac{1}{1+s^2}\space\text{d}s-2\int\frac{1}{1+2s^2}\space\text{d}s$$
So, the answers are:
*
*$$\text{I}_1=\frac{\arctan\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}\approx0.43521$$
*$$\text{I}_2=\sqrt{2}\arctan\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}\approx0.0850216$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| By Vieta's formulas $x$ and $y$ are roots of the polynomial
$$ p(z)=z^2-3z+3 $$
and if $z\in\{x,y\}$ we have $z^2=3z-3$, from which $z^4=9z^2-18z+9=9z-18$ and
$$ x^4+y^4 = 9(x+y)-36 = \color{red}{-9}.$$
In general, the characteristic polynomial of the sequence $\{s_n=x^n+y^n\}_{n\geq 0}$ is exactly $p(z)$,
so $s_{n+2}=3(s_{n+1}-s_n)$. Since $s_0=2$ and $s_1=3$, it follows that $s_2=3(3-2)=3$ and
$s_3 = 3(3-3)=0$, so $s_4=3(0-3)=-9$ and $s_5=3(-9-0)=-27$, for instance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Proof Verification : Prove -(-a)=a using only ordered field axioms I need to prove for all real numbers $a$, $-(-a) = a$ using only the following axioms:
Thanks to many members of the Mathematics Stackexchange Community, I have the following proof worked out:
Theorem: The Additive Inverse Identity is Unique
$
( \forall a,b,c \in \mathbb{R} )(a+b=0) \land (a+c=0) \\
( \forall a,b,c \in \mathbb{R} )(a+b=a+c) \\
( \forall a,b,c \in \mathbb{R} )(b=c) \\
\text{QED} \\
$
Theorem $\textbf{a} \cdot \textbf{0 = 0}$
$
\begin{align}
a \cdot 0 &= a \cdot 0 \\
a \cdot (0 + 0) &= a \cdot 0 \\
a \cdot 0 + a \cdot 0 &= a \cdot 0 \\
a \cdot 0 + a \cdot 0 + (-a) \cdot 0 &= a \cdot 0 + (-a) \cdot 0 \\
a \cdot 0 &= 0 \\
&\text{QED} \\
\end{align}
$
Theorem: $-\textbf{1} \cdot \textbf{(a) = (}-\textbf{a)}$
$
\begin{align}
a \cdot 0 &= 0 \\
a \cdot \left[ 1 + (-1) \right] &= a + (-a) \\
1 \cdot a + (-1) \cdot a &= a + (-a) \\
-1 \cdot (a) &= (-a) \\
&\text{QED} \\
\end{align}
$
Theorem: $-\textbf{(}-\textbf{a) = a}$
$
\begin{align}
0 &= 0 \\
-a \cdot 0 &= 0 \\
-a \cdot \left[ 1 + (-1) \right] &= 0 \\
-a \cdot 1 + -a \cdot (-1) &= 0 \\
-a + \left[ -(-a) \right] &= 0 \\
-(-a) &= a \\
\text{QED} \\
\end{align}
$
Does everything look alright? Have I missed anything?
Also--why is it necessary to show that the additive inverse is unique?
Many thanks in advance.
| An extended comment:
*
*If you want a proof verification it make sense that you number your equations so that they are easy to reference. You can use \$\tag{1}\$ in the equation code and reference it as \$(1)\$.
*Start end end your LaTeX blocks wiht \$\$ and not with \$.
Lets make
$$
\begin{align}
a \cdot 0 + a \cdot 0 &= a \cdot 0 \tag{1.1}\\
a \cdot 0 + a \cdot 0 + (-a) \cdot 0 &= a \cdot 0 + (-a) \cdot 0 \tag{1.2}\\
a \cdot 0 &= 0 \tag{1.3}\\
\end{align}
$$
more precise, then you have
$$
\begin{align}
a \cdot 0 + a \cdot 0 &= a \cdot 0 \tag{2.1}\\
(a \cdot 0 + a \cdot 0 )+ (-a) \cdot 0 &= a \cdot 0 + (-a) \cdot 0 &\text{(you have to use parantheses)}\tag{2.2}\\
a \cdot 0 + (a \cdot 0 + (-a) \cdot 0) &= a \cdot 0 + (-a) \cdot 0 & \text{(by associativity of +)} \tag{2.3}\\
a \cdot 0 + ((a+(-a)) \cdot 0 &= ((a+(-a)) \cdot 0 &\text{(by distributive law)}\tag{2.4}\\
a \cdot 0 + 0 \cdot 0 &= 0 \cdot 0 &\text{(by law of inverse)}\tag{2.5}\\
a \cdot 0 +0&= 0 & \text{(because }0\cdot0=0 \text{)}\tag{2.6}\\
a \cdot 0 &= 0& \text{0 is identity element of +} \tag{2.7}\\
\end{align}
$$
To get $(6)$ you use still an unproven identity:
$$0\cdot 0=0 \tag{3.1}$$
It is simpler you proceed in the following way
$$
\begin{align}
a \cdot 0 + a \cdot 0 &= a \cdot 0 \tag{4.1}\\
(a \cdot 0 + a \cdot 0) + (-( a \cdot 0)) &= a \cdot 0 + (-( a \cdot 0)) &\text{(you have to use parantheses)}\tag{4.2}\\
a \cdot 0 + (a \cdot 0 + (-( a \cdot 0))) &= a \cdot 0 + (-( a \cdot 0)) & \text{(by associativity of +)}\tag{4.3}\\
a \cdot 0 + 0 &= 0 & \text{(inverse element)}\tag{4.4}\\
a \cdot 0 &= 0& \text{(0 is identity element of +)} \tag{4.5}\\
\end{align}
$$
The latter proof is shorter and complete. The first proof lacks the proof of
$(3.1)$.
Edit
I forgot the most important part of my answer.
You showed that in a field with operations + and $\cdot$ we have
$$-(-a)=a$$
by using the distributive law.
But we have
$$
\begin{align}
a+(-a)&=0 &\text{right inverse of }a\tag{5.1}\\
(-(-a))+(-a)&=0& \text{left inverse of }(-a) \tag{5.2}\\
a+(-a)&=(-(-a))+(-a) &\tag{5.3}\\
(a+(-a))+a&=((-(-a))+(-a))+a & \tag{5.4}\\
a+((-a)+a)&=-(-a)+((-a)+a) & \text{associativity}\tag{5.5}\\
a+0&=-(-a)+0 & \text{left inverse}\tag{5.6} \\
a &= -(-a) & \text{identity}\tag{5.7} \\
\end{align}$$
So you do not need a $\cdot$ operation to show this property of the inverse of th $+$ operator. We can say therefore
If $(G,+)$ is a group and $-g$ the inverse of $g \in G$ then $-(-g)=g$
And from this immediately follows
If $(G,\cdot)$ is a group and $g^{-1}$ the inverse of $g \in G$ then $(g^{-1})^{-1}=g$
So from your Axioms follows:
$$-(-a))=a,\; (a^{-1})^{-1}=a\tag{6.1}$$
$(6.1)$ already follows from P1,P2,P3,P5,P6,P7
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Proof related to Fibonacci sequence Could anyone help me with this problem?
$$
\sum_{j=0}^{n}\binom{n}{j}F_{n+1-j}= F_{2n+1}
$$
I used induction and was able to get to this:
$$
2\sum_{j=0}^{n}\binom{n}{j}F_{n-j}= F_{2n}
$$
However I still do not know how to prove the second equality. Really appreciate any help
| (For future reference as an exercise in integration.) Suppose we seek
to verify that
$$F_{2n} = \sum_{j=0}^n {n\choose j} F_{n-j}$$
with $F_q$ being a Fibonacci number. These have generating function
$$f(z) = \frac{z}{1-z-z^2}$$
and hence
$$F_{n-j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-j+1}} \frac{z}{1-z-z^2} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{z}{1-z-z^2}
\sum_{j=0}^n {n\choose j} z^j
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{z}{1-z-z^2}
(1+z)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} \frac{1}{1-z-z^2}
(1+z)^n
\; dz.$$
Now put $z/(1+z) = v$ so that $z = v/(1-v)$ and $dz = 1/(1-v)^2 dv$ to
get
$$\frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{n}}
\frac{1}{1-v/(1-v)-v^2/(1-v)^2}
\frac{1}{(1-v)^2}
\; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{n}}
\frac{1}{(1-v)^2-v(1-v)-v^2}
\; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{n}}
\frac{1}{1-2v+v^2-v+v^2-v^2}
\; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{n}}
\frac{1}{1-3v+v^2}
\; dv.$$
This is
$$[v^n] \frac{v}{1-3v+v^2}.$$
Observe however that the generating function of the even index
Fibonacci numbers is given by
$$\left.\left(\frac{1}{2} f(z) + \frac{1}{2} f(-z)
\right)\right|_{z^2=w}.$$
We thus have for the generating function
$$\left. \left(\frac{1}{2} \frac{z}{1-z-z^2}
- \frac{1}{2} \frac{z}{1+z-z^2}\right)\right|_{z^2=w}
\\ = \left. \frac{1/2z(1+z-z^2)-1/2z(1-z-z^2)}{(1-z^2-z)(1-z^2+z)}
\right|_{z^2=w}
\\ = \left. \frac{z^2}{(1-z^2)^2-z^2}
\right|_{z^2=w}
= \frac{w}{1-3w+w^2}$$
and the claim is established.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of trigonometric equation $\cos^2(\cos\theta) + \sin^2(\sin\theta)$ For any real $\theta$ the maximum value of $$\cos^2(\cos\theta) + \sin^2(\sin\theta)$$
A. $1$
B. $1 + \sin^21$
C. $1 + \cos^21$
D. does not exist
I tried it by converting the whole expression into $\sin$ but getting nowhere with that.
$$1-\sin^2(\cos\theta) + \sin^2(\sin\theta)$$
Now since 1 is constant therefore, $$[\sin^2(\cos\theta) + \sin^2(\sin\theta)]$$ should be minimum but I don't know how to minimize it.
Also is there a way to think about it's solution graph.
I have to solve this without using calculus.
Kindly help.
| $$\max_{\theta\in\mathbb{R}}\{\cos^2(\cos\theta) + \sin^2(\sin\theta)\}\le\max_{-1\le x \le 1}\cos^2 x+\max_{-1\le y \le 1}\sin^2 y=1+\sin^21$$
$$\cos^2 \left( \cos \frac{\pi}{2} \right) + \sin^2 \left( \sin \frac{\pi}{2} \right) = 1 + \sin^2 1$$
So, the maximum is at most $1+\sin^21$, and this value is achieved. Hence the answer is b).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to calculate this imaginary part of a complex square root? The calculation of the square root of a complex number $a + ib$ involves solving the equation
$$ (x + iy)^2 = a + ib$$
So far so good. One obtains the equations
$$ 4x^4 -4ax^2 - b^2 = 0, y = b/2x$$
and using the quadratic formula for $x^2$ one gets
$$ x = \pm \sqrt{{a + \sqrt{a^2 + b^2} \over 2}}$$
I am supposed to get
$$ y = \pm \sqrt{- a + \sqrt{a^2 + b^2} \over 2} \cdot \text{sgn}(b)$$
but if I substitute $x$ into $y$ I get
$$ y = {b\over 2x} = \pm {b \sqrt{2} \over 2\sqrt{a + \sqrt{a^2 + b^2}}}$$
What am I doing wrong?
| *
*When $\text{a}\space\wedge\space\text{b}\in\mathbb{C}$:
$$\text{a}=\text{b}^2=\left(\Re[\text{b}]+\Im[\text{b}]i\right)^2=\Re^2[\text{b}]-\Im^2[\text{b}]+2\Re[\text{b}]\Im[\text{b}i]$$
*Solving $x$:
$$y=\frac{\text{b}}{2x}\Longleftrightarrow\text{b}=2xy\Longleftrightarrow x=\frac{\text{b}}{2y}$$
*Solving $x$ by substituting $y=x^2$:
$$4x^4-4\text{a}x^2-\text{b}^2=0\Longleftrightarrow y=\frac{\text{a}\pm\sqrt{\text{a}^2+\text{b}^2}}{2}\Longleftrightarrow x=\color{red}{\pm}\frac{\sqrt{\text{a}\color{red}{\pm}\sqrt{\text{a}^2+\text{b}^2}}}{\sqrt{2}}$$
So for (3) we find 4 solutions.
And for $\text{sgn}(\text{z})$, when $\text{z}\in\mathbb{C}$ (and $\text{z}\ne0$):
$$\text{sgn}(\text{z})=\frac{\text{z}}{\sqrt{\text{z}\cdot\overline{\text{z}}}}=\frac{\text{z}}{\sqrt{|\text{z}|^2}}=\frac{\text{z}}{|\text{z}|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculus 2: Partial fractions problem. Finding the value of a constant I encountered the following problem.
Let $f(x)$ be a quadratic function such that $f(0) = -6$ and
$$\int \frac{f(x)}{x^2(x-3)^8} dx $$
is a rational function.
Determine the value of $f'(0)$
Here's what I tried. I decomposed the fraction integrand below
$$\frac{f(x)}{x^2(x-3)^8} = \frac{A_1}{x} + \frac{A_2}{x^2} +\sum_{i=1}^8 \frac{B_i}{(x-3)^i}$$
By finding a common denominator, I determined
$$f(x) = A_1x(x-3)^8 + A_2(x-3)^8 +x^2 \sum_{j=1}^8 [B_j(x-3)^{8-j} ]$$
$$f'(x) = A_1(x-3)^8 + 8A_2(x-3)^7 + D(x) $$
where $D(x)$ is a function such that $D(0) = 0$. (These are all the remaining terms that go away when we plug $0$ into $f'(0)$).
I used the information that $f(0) = -6$ to get the equation
$$f(0) = A_2(-3)^8 = -6 \rightarrow A_2 = \frac{-2}{3^7}$$.
This leaves us with
$$f'(0) = A_1(-3)^8 +8\cdot \frac{-2}{3^7} \cdot (-3)^7 $$
$$ f'(0)= 3^8 \cdot A_1 + 16$$
I was told $f'(0) = 16$, however I cannot convince myself the value of $A_1$. I would think that the fact that $f(x)$ is a quadratic function should come into play here. Please let me know what you think.
| Note that since $f$ is quadratic with $f(0)=0$, we can write $f(x)$ as $$f(x)=-6+f'(0)x+\frac12 f''(0)x^2$$
Then, the integrand becomes
$$\frac{f(x)}{x^2(x-3)^8}=\frac{-6}{x^2(x-3)^8}+\frac{f'(0)}{x(x-3)^8}+\frac{\frac12 f''(0)}{(x-3)^8} \tag 1$$
The last term on the right-hand side of $(1)$ integrates to the rational function, $\int \frac{\frac12 f''(0)}{(x-3)^8}\,dx =-\frac12f''(0)\frac{1}{7(x-3)^7}+C$ and is not implicated , therefore, in the ensuing analysis.
Using partial fraction expansion, we can write the first term on the right-hand side of $(1)$ as
$$\frac{-6}{x^2(x-3)^8}=\frac{-16}{6561\,x}+\frac{16}{6561\,(x-3)}+\left(\text{other terms that integrate to rational functions}\right) \tag 2$$
and the second term on the right-hand side of $(1)$ as
$$\frac{f'(0)}{x(x-3)^8}=\frac{f'(0)}{6561\,x}-\frac{f'(0)}{6561\,(x-3)}+\left(\text{other terms that integrate to rational functions}\right) \tag 3$$
Therefore, to annihilate the terms that integrate to $\log(x)$ and $\log(x-3)$ in $(2)$ and $(3)$, we must have $$f'(0)=16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $ My answer: $2^\frac{-1}{\sqrt{2}} < x < 1$
Textbook answer: $2^\frac{-12}{17} < x < 1$
The only difference between my resolution and the Textbook one is that I solved by saying that
$$\text{(I) }\sqrt{1 - 8\cdot (\log_{1/4}{x})^2} > 1 -3\cdot \log_{1/4}x $$
Is true when II OR III are true:
$$\text{(II) } 1 - 3\cdot \log_{1/4}x > 0 \quad \land \quad 1-8\cdot (\log_{1/4}{x})^2 > (1 - 3\cdot \log_{1/4}x)^2 $$
$$\text{(III) } 1 - 3\cdot\log_{1/4}x < 0 \quad \land \quad 1- 8\cdot(\log_{1/4}{x})^2 > 0 $$
And the Textbook said that (I) is true when:
$$ \text{(IV) }1- 8\cdot(\log_{1/4}{x})^2 > (1 - 3\cdot \log_{1/4}x)^2 \ge 0 $$
Sorry if this is a dumb question, any help is appreciated. I can't really understand if my conditions (II or III) and their condition (IV) are both right or if one of them is wrong.
| at first we have $$x>0$$ and $$\frac{1}{8}\geq \frac{(\ln(x))^2}{(2\ln(2))^2}$$
this condition gives us
$$e^{\ln(2)/\sqrt{2}}\geq x\geq1$$
or
$$e^{-\ln(2)/\sqrt{2}}\le x<1$$
now we can solve the inequality
$$1+\frac{3\ln(x)}{2\ln(2)}<\sqrt{\left(1-\frac{2\ln(x)^2}{\ln(2)^2}\right)}$$
and we do case work:
I)if $$x\le e^{-\frac{2\ln(2)}{3}}$$ is hold then our inequality is true.
II) if $$x> e^{-\frac{2\ln(2)}{3}}$$ is hold then we get after squaring and simplifying
$$\frac{17}{4}(\ln(x))^2+3\ln(2)\ln(x)<0$$
from the last inequality we obtain
$$e^{-12/17\ln(2)}<x<1$$
with the condition above we get finaly $$e^{-2\ln(2)/2}<x<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim
Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$
For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt 3} = 768614336404564651$ which is prime. Then $\binom{2^{61}}{2}-1 = 2658455991569831744654692615953842175$ and it's largest prime factor is $q$. Note ${2^{29}+1 \above 1.5pt 3} =178956971$ is not prime and the largest prime factor of $\binom{2^{29}}{2}-1$ is $3033169$. But even then we could note that $178956971 = 59*3033169$. A counterexample to the claim would suffice to to show it's wrong.
| Since you already have other answers, I will instead offer a small generalization.
Suppose $p=\frac{k^n +1}{k+1}$ is a prime number. Then, $p$ is the largest prime factor of $\binom{k^n}{2}-1$ if $k=2j^2$ for some integer $j$.
Proof:
We have:
$$\binom{k^n}{2}-1 = \frac{1}{2}k^n(k^n-1)-1 = \frac{1}{2}(k^{2n}-1)-\frac{1}{2}(k^n+1) = \frac{1}{2}(k^n+1)(k^n -2)$$
This can of course be rewritten as $\frac{k+1}{2}p (k^n-2)$
Now suppose $k=2j^2$ for some $j \in \mathbb{N}$. Then, note that since $k \equiv -1 \mod (k+1)$, $n$ must be an odd integer in order for $\frac{k^n +1}{k+1}$ to have integer values. Thus, $\frac{n-1}{2}$ is also an integer. Then:
$$\frac{k+1}{2}p (k^n-2) = (k+1)p (2^{n-1}j^{2n}-1) = (k+1)p(2^{\frac{n-1}{2}}j^{n}-1)(2^{\frac{n-1}{2}}j^{n}+1)$$
And certainly we have that all the other factors are less than $p$, so this is the largest prime factor. (Note: this can be seen because the other 3 factors $\neq p$ in $(k+1)p(2^{\frac{n-1}{2}}j^{n}-1)(2^{\frac{n-1}{2}}j^{n}+1)$ must all be less than $p$, regardless of whether or not they are prime, implying $p$ has to be the largest prime factor).
EDIT: At the request of the person who asked the original question, I will provide a proof of the following (a quick counterexample to the contrary statement is $n=81$).
Suppose $k$ arbitrary. Then it is always possible to find $n$ such that $3^k$ divides $\binom{2^n}{2}-1$. In particular, $n = 3^{k-2}$ always works.
Proof:
This proof will use the Lifting the Exponent (LTE) Lemma. I suggest looking up the statement if you want to fully understand the proof.
Let $k$ be given. Then we can rewrite $\binom{2^n}{2}-1 = (2^n+1)(2^{n-1}-1)$. In order for either one of these factors to be divisible by $3$, we must have that $n$ is an odd number (since $(-1)^n \equiv -1 \mod 3$). Then $n-1 = 2j$, an even integer. We can then rewrite the above:
$$\binom{2^n}{2}-1 = (2^n+1)(2^{n-1}-1) = (2^n+1)(4^{j}-1)$$
We can now apply the LTE Lemma. Define a valuation $v$ as follows: $v_p (x) = y$ implies that $y$ is the largest power of $p$ that divides $x$, where $p$ is prime. Then LTE Lemma says:
$$v_3((2^n+1)(4^{j}-1)) = v_3(2^n+1) + v_3(4^{j}-1) = v_3(3) + v_3(n)+v_3(3)+v_3(j)$$
Note that $v_3(3) = 1$ of course. Then, note $j = (n-1)/2$, we have:
$$v_3((2^n+1)(4^{j}-1)) = 2 + v_3 \Big( \frac{n(n-1)}{2} \Big)$$
Now, in the above, note that either $n$ or $n-1$ can be divisible by $3$, but not both, so we can assume without loss of generality that $n-1$ is not divisible be $3$, ie $v_3(n-1) = 0$. Also, since a valuation is logarithmic, the $2$ in the denominator goes away since $v_3(2) = 0$. Thus we have the following:
$$v_3((2^n+1)(4^{j}-1)) = 2+v_3(n)$$
Then we are done. Suppose $k$ is given. Then firstly, we have that $3^2 = 9$ will always divide $\binom{2^n}{2}-1$. Now suppose we want $3^k$ to divide it. By the above, we only need to choose $n = 3^{k-2}$ (or any multiple of this).
Actually ... you can generalize this to the $\binom{k^n}{2}-1$ case, but don't worry I will spare you the details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Finding the value of $x+y+z$ given sums of two and square root of the other Given: $$x+y+\sqrt{z}=148$$
$$x+\sqrt{y}+z=82$$
$$\sqrt{x}+y+z=98$$
Find the value of $x+y+z$ such that x, y, and z are positive integers.
The only significant step I did was to let $a^2=x, b^2=y, c^2=z.$ After that, I just played with the equations, adding and subtracting, letting $S=x+y+z$, finding equations for S. But I always come to a dead end. Can anyone show me the solution?
| Following on from your step, we subtract the second equation from the third. We then have
$$a-a^2 +b^2-b =16$$
or
$$(b-a)(b+a-1)=16$$
No the possible combinations are: $\{1,16\}$, $\{2,8\}$, or $\{4,4\}$. A quick check tells you only the first will work. Thus $b-a=1$ and $b+a-1= 16$. From this we get $a=8$ and $b=9$. Subtracting the third equation from the first we have
$$
a^2-a + c -c^2=50$$ from which we can solve for $c$ and we obtain $c(c-1)=6$ or $c=3$. Thus $x=64, y =81, z =9.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| We have
$$a^4\le a^4+b^4+c^4+d^4=16\implies a\le 2$$
So, we can have $a^4(a-2)\le 0$, i.e.
$$a^5\le 2a^4$$
Similarly,
$$b^5\le 2b^4,\quad c^5\le 2c^4,\quad d^5\le 2d^4$$
giving
$$a^5+b^5+c^5+d^5\le 2(a^4+b^4+c^4+d^4)=32$$
Note here that
$$a^5+b^5+c^5+d^5=2(a^4+b^4+c^4+d^4)$$
holds when
$$a^4(a-2)=b^4(b-2)=c^4(c-2)=d^4(d-2)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Intermediate Modular Arithmetic Strategy And Questions I have these problems and I know how to solve ones with two equations but not more. Can anyone provide solutions? Thanks in advance!
Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 1 \pmod{7}, \\
N &\equiv 7 \pmod{13}, \\
N &\equiv 13 \pmod{20}.
\end{align*}
Find the smallest positive $N$ such that
\begin{align*}
N &\equiv 6 \pmod{12}, \\
N &\equiv 6 \pmod{18}, \\
N &\equiv 6 \pmod{24}, \\
N &\equiv 6 \pmod{30}, \\
N &\equiv 6 \pmod{60}.
\end{align*}
How many positive integers less than or equal to $6\cdot7\cdot8\cdot9$ solve the system of congruences
\begin{align*}
m &\equiv 5 \pmod{6}, \\
m &\equiv 4 \pmod{7}, \\
m &\equiv 3 \pmod{8}, \\
m &\equiv 3 \pmod{9}.
\end{align*}
| For problem 1:
$$N=7p+1$$
$$7p+1 \equiv 7 \pmod{13}$$
$$7p \equiv 6 \pmod{13}$$
$$p \equiv 12 \pmod{13}$$
(This is calculated by finding the inverse of 7 mod 13).
$$N \equiv 7*12+1 \pmod{13*7}$$
$$N \equiv 85 \pmod{91}$$
$$N = 91q + 85$$
$$N \equiv 13 \pmod{20} $$
$$91q + 85 \equiv 13 \pmod{20} $$
$$91q \equiv 8 \pmod{20} $$
$$11q \equiv 8 \pmod{20}$$
$$q \equiv 8 \pmod{20} $$
$$N \equiv 91*8+85 \pmod{91*20}$$
$$N \equiv 813 mod \pmod{1820}$$
$$N = 813$$
For problem 2, $N = 6$ is the obvious solution.
For problem 3, we see that it is not possible to have $N \equiv 5 \pmod{6}$ and $N \equiv 3 \pmod{9}$ (first suggests $N$ is not divisible by 3, second suggests $N$ is divisible by 3). So, here there are 0 valid solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of Trigonometric Integral
Evaluation of $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx$
$\bf{My\; Try::} $ Let $$I= \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{(\sin x+\cos x)(2-\sin 2x)}dx$$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{\sin\left(x+\frac{\pi}{4}\right)(2-\sin 2x)}dx$$
Put $\displaystyle x+\frac{\pi}{4} = t,$ Then $dx = dt$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t}{\sin t(2+\cos 2t)}dt=\frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-4+4}{\sin t(2+\cos 2t)}dt$$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-2}{\sin t}dt+\frac{2}{\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{1}{\sin t(2+\cos 2t)}dt$$
How can i solve it after that, Help required, Thanks
| Hint
From where you left off:
The first integrand is the same as $$-\csc t-4\sin t+4\sin^3t$$
The second integral will be evaluated using the substitution $u=\cos t$ followed by a dose of partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the probability of the event that $c=a+b$
In a set $\{0,1,2,3,\dots,n\}$, $3$ numbers $(a,b,c)$ are randomly chosen.
What is the probability of the event that $c=a+b$?
I thought, I should start with $c=0$ for which there is only $1$ way $c=0$, if both $a$ and $b$ are $0$.
For $c=1, 2$ ways: $a=1$ and $b=0$ or $a=0$ and $b=1$.
For $c=2, 3$ ways: $a=2$ and $b=0, a=0$ and $b=2$, $a=1$ and $b=1$,
and so on.
If we follow the pattern we can see that the number of outcomes for the terms provided is $c+1$.
I am stuck at the total number of outcomes. Since the set is from $0$ to $n$, logically the probability that $c=a+b$ should be $0$.
How can I continue this?
Thank you
| You pointed out that given $c$, the number of ways to have $a+b=c$ is $c+1$. We now must figure out how many total ways there are to have $a+b=c$. This is a simple sum over the range of $c$:
$$\sum_{c=0}^n c+1$$
$$n+1 +\sum_{c=0}^n c$$
$$n+1 +\frac{n(n+1)}{2}$$
$$\frac{(n+2)(n+1)}{2}$$
To find the probability of this event occurring, we need only divide by the total number of possible outcomes, which is $(n+1)^3$:
$$\frac{(n+2)(n+1)}{2(n+1)^3}$$
$$\frac{(n+2)}{2(n+1)^2}$$
which is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the sum of all positive even divisors of 1000? I know similar questions and answers have been posted here, but I don't understand the answers. Can anyone show me how to solve this problem in a simple way? This is a math problem for 8th grade students.Thank you very much!
What is the sum of all positive even divisors of 1000?
| $n$ is a positive even divisor of $1000$ if and only if $n = 2m$ where $m$ is a divisor of $500$. Since $500 = 2^2 \times 5^3$, there are $(2+1)(3+1) = 12$ divisors of $500$. Those divisors are
\begin{array}{rr}
1, & 500, \\
2, & 250, \\
4, & 125, \\
5, & 100, \\
10, & 50, \\
20, & 25 \\
\end{array}
so there are $12$ positive even divisors of $1000$.
Those divisors are
\begin{array}{cc}
2, & 1000, \\
4, & 500, \\
8, & 250, \\
10, & 200, \\
20, & 100, \\
40, & 50 \\
\end{array}
The sum of the positive divisors of $500 = 2^2 \times 5^3$ equals
$\dfrac{2^3 - 1}{2 - 1} \times \dfrac{5^4 - 1}{5 - 1} = 1092$
So the sum of the even divisors of $1000$ is $2 \times 1092 = 2184$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 6,
"answer_id": 2
} |
Calculus - Finding limit (NOT L'Hopital's Rule): $\lim_{x \to 1^-}\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}$ How do I find this limit?
$$\displaystyle{\lim_{x \to 1^-}}\frac{x^2+x+\sin({\pi \over 2}x)-3}{x-1}$$
I am unable to factor the numerator to get rid of the denominator. Can someone please help? Thank you!
Is there any other way to get the answer besides using L'Hopital's Rule?
| Another possible way to do it.
Start changing variable $x=y+1$; this gives
$$\frac{x^2+x+\sin \left(\frac{\pi x}{2}\right)-3}{x-1}=\frac{y^2+3 y+\cos \left(\frac{\pi y}{2}\right)-1}{y}$$ Now, use Taylor expansion $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ which gives $$\cos \left(\frac{\pi y}{2}\right)=1-\frac{\pi ^2 y^2}{8}+O\left(y^4\right)$$ So $$\frac{y^2+3 y+\cos \left(\frac{\pi y}{2}\right)-1}{y}=\frac{y^2 +3y-\frac{\pi ^2 y^2}{8}+O\left(y^4\right)}{y}=(1-\frac{\pi ^2 }{8})y+3+O\left(y^3\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
When does this "primes" matrix have full row and column rank? When does the following type of matrix have full row and column rank?
For each entry, we take the value $v$=row*column.
Then for the $b$th bit of $v$, $v_b$ we take the $b$th prime to the $v_b$ power.
EXAMPLES
Say row = 5 and column = 7.
This is 5*7=35, which, in binary, is $100011_2$
So we take
$$\cdot 13^1 \cdot 11^0 \cdot 7^0 \cdot 5^0 \cdot 3^1 \cdot 2^1$$
$$=2 \cdot 3 \cdot 13$$
$$=78$$
...So the entry for the 5th row and 7th column would be 78.
Say row = 4 and column = 2.
This is 4*2=8, which, in binary, is $001000_2$
So we take
$$\cdot 13^0 \cdot 11^0 \cdot 7^1 \cdot 5^0 \cdot 3^0 \cdot 2^0$$
$$=7$$
...So the entry for the 4th row and 2nd column would be 7.
THE GENERAL MATRIX
$$
\begin{matrix}
2^0 3^0 5^0 & 2^0 3^0 5^0 & 2^0 3^0 5^0 & 2^0 3^0 5^0 & 2^0 3^0 5^0 & \dots\\
2^0 3^0 5^0 & 2^1 3^0 5^0 & 2^0 3^1 5^0 & 2^1 3^1 5^0 & 2^0 3^0 5^1 & \dots\\
2^0 3^0 5^0 & 2^0 3^1 5^0 & 2^0 3^0 5^1 & 2^0 3^1 5^1 & 2^0 3^0 5^0 7^1 & \dots\\
2^0 3^0 5^0 & 2^1 3^1 5^0 & 2^1 3^1 5^0 & 2^1 3^0 5^0 7^1 & 2^0 3^0 5^1 7^1 & \dots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots\\
\\
\end{matrix}
$$
We can assume that the matrix is an $N \times N$ matrix.
SOME OBSERVATIONS
Just a general observation: The only counterexample for $N \times N$ matrices with $N$ ranging from 2 to 20 is $N=3$. All other matrices have full rank.
| Easy counterexample, take $N = 3$ and you will see that it is not full ranked:
$\left( \begin{array}{ccc}
2^03^05^0=1 & 2^03^05^0=1 & 2^03^05^0=1 \\
2^03^05^0=1 & 2^13^05^0=2 & 2^03^15^0=3 \\
2^03^05^0=1 & 2^03^15^0=3 & 2^03^05^1=5 \end{array} \right)$, this matrix is not full rank, so your question's answer would be no, it is not always full rank.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - 1) = 2(\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}})$$
And as I solved it, the equation came down to:
$2\operatorname{cosec}^{2}{B}+2\cot^{2}{B}.$
This can be written as $2(\operatorname{cosec}^{2}{B} + \cot^{2}{B}),$ which can further be written as $2\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}}.$
But I can't seem to get my mind around the middle part of the question, that is, $2(\operatorname{cosec}^{2}{B} - 1)$
Is it possible to write it like this, or is this an error in the question paper itself?
| For the first part-
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} $$
$$=\frac{(\operatorname{cosec}B-\cot B)^2+(\operatorname{cosec}B-\cot B)^2}{1}$$
$$=2(\operatorname{cosec^2}B+\cot^2B)$$
$$=2(2\operatorname{cosec^2}B-1)$$
For the second part $$2(\frac{1+\cos^2B}{1-\cos^2B})=2(\frac{1+\cos^2B}{\sin^2B})=2(\operatorname{cosec}^2B+\cot^2B)$$.Now,proceed as previous part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality:
$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$
I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\left(-4+3 \sin^2 \theta\right)$ from the right side, but I don't know where to go from there.
| Let $y=2\sin2x+3\cos^2x=\cos^2x(4\tan x+3)$
$$\iff y\tan^2 x-4\tan x+y-3=0$$ which is a Quadratic Equation in $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
i.e., $$4^2-4y(y-3)=-4(y+1)(y-4)\ge0\iff(y+1)(y-4)\le0$$
$$\iff-1\le y\le4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding $x$ such that $2x^2-2x+1=(2m+1)^2$ We need to find the smallest integral value of $x$ such that $2x^2-2x+1=(2m+1)^2$ and $x \ge 10^{15}$.
| According to Stefan4024, the solution for the Pythagorean triples are well-known:
$$\begin{pmatrix} x-1 \\ x \\ 2m+1 \end{pmatrix}=
\begin{pmatrix}
\frac{(\sqrt{2}+1)^{2k+1}-(\sqrt{2}-1)^{2k+1}}{4}-\frac{1}{2} \\
\frac{(\sqrt{2}+1)^{2k+1}-(\sqrt{2}-1)^{2k+1}}{4}+\frac{1}{2} \\
\frac{(\sqrt{2}+1)^{2k+1}+(\sqrt{2}-1)^{2k+1}}{2\sqrt{2}}
\end{pmatrix}$$
Now $\dfrac{(\sqrt{2}+1)^{2k+1}}{4} \approx 10^{15} \implies 2k+1\gtrsim 40.76$
Put $2k+1=41$, then
$$\begin{pmatrix} x-1 \\ x \\ 2m+1 \end{pmatrix}=
\begin{pmatrix}
1235216565974040 \\
1235216565974041 \\
1746860020068409
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding limit of sequence, done right: $a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$? I need to find the limit of this sequence:
$a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$
So I multiply with this since $(a-b)(a+b)=(a^2-b^2)$
$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$
And get $\dfrac{9-n+9}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$
Then divide by n?
And get $\dfrac{9/n-n/n+9/n}{\sqrt{n^2/n+9/n}+\sqrt{n^2/n-n+9/n}}$
$\lim = \dfrac{1}{\infty} = 0$?
| After multiplying the initial equation by $$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$$
You should get $(n^2+9) - (n^2 - n +9) = n$ in the numerator, i.e., you should have this:
$$\frac{n}{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}$$
Now try dividing numerator and denominator by $n$:
$$\lim_{n\to \infty} \frac n{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}= \lim_{n\to \infty} \frac 1{\sqrt{1+\frac 9{n^2}}
+\sqrt{1-\frac 1n+\frac 9{n^2}}} = \frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $ Given the followning limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $$
To find limit I have made following steps:
*
*Let $ x = y $ ,then limit equals $0$
*Let $ x > y $ ,then consider the limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{x^2}{x^2 + y^2}\right)^{x^2} = \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{1}{1 + \frac{y^2}{x^2}}\right)^{x^2} = 0$$
with respect to $$0 < y^2/x^2 < const$$
*Let $ y > x $ ,then consider the limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{x^2}{x^2 + y^2}\right)^{y^2} = \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{1}{\frac{x^2}{y^2} + 1}\right)^{x^2} = 0$$
with respect to $$0 < x^2/y^2 < const$$
What could you say about my solution?
| Hint,
Assume $x = r \cos \theta$ and $y = r \sin \theta, \theta = constant$
Then, the limit changes to
$$ \lim \limits_{r \rightarrow \infty}\left ( \dfrac {r^2 \sin \theta \cos \theta}{r^2}\right)^{r^2 \cos^2 \theta} \\ \Rightarrow \lim \limits_{r \rightarrow \infty} (\sin \theta\cos \theta)^{r^2 \cos^2\theta} = 0$$
As expression under bracket is $<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Quadric obtained by rotation of a line I have a line of equations:
\begin{align}r:
\begin{cases}
x = 2+2t \\
y = 2 - t \\
z=t
\end{cases}
\end{align}
and another line of equations:
\begin{align}s:
\begin{cases}
x = 3 \\
y = 1 \\
z=k
\end{cases}
\end{align}
I have to determine what quadric I get when i rotate line $r$ about $s$.
Every single point of the line that rotates creates a circumference of equation
$x^2+y^2=r^2$
hence all those infinte points creates a quadric.
What is the best and fastest way to get to the result?
| We have line of equation
\begin{align}r:
\begin{cases}
x=f(t)=2+2t\\
y=g(t)=2-t\\
z=h(t)=t
\end{cases}
\end{align}
and the axis of the rotation
\begin{align}s:
\begin{cases}
x=x_0+at=3\\
y=y_0+bt=1\\
z=z_0+ct=k
\end{cases}
\end{align}
Let's choose a generic point on line $s$ and let it be $C(3,1,0)$.
We have to find the plane through $C$ and perpendiular to $r$.
The plane has generic equations
$a(x-f(t))+b(y-g(t))+c(z-h(t))=0$
So our plane has the same parameters as $s$ $(0,0,1)$ so our plane is
$z-t=0$, in particular $t=z$.
Now we need to find the sphere with centre in $P(3,1,0)$ and radius $\overline{CP}$ knowing that $P$ is a point on line $r$ and has equations $P(f(t), g(t), h(t))$.
The generic sphere has equation:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=(f(t)-x_0)^2+(g(t)-y_0)^2+(h(t)-z_0)^2$
Then our equations becomes
$(x-3)^2+(y-1)^2 +(z-0)^2 = (2+2t-3)^2+(2-t-1)^2+(t-0)^2$
$x^2-6x+9+y^2-2y+1+z^2 = (2t-1)^2+(1-t)^2+t^2$
$x^2-6x+9+y^2-2y+1+z^2 = 4t^2-4t+1+1+t^2-2t+t^2$
$x^2-6x+9+y^2-2y+1+z^2 = 4z^2-4z+1+1+z^2-2z+z^2$
And the equation that we get is
$x^2+y^2-5z^2-6x-2y+6z+8=0$
It is an hyperboloid as we can verifiy with the matrices
\begin{align}I_3=
\begin{vmatrix}
1 & 0 & 0 & -6 \\
0 & 1 & 0 & -2 \\
0 & 0 & -5 & 6 \\
-6 & -2 & 6 & 8
\end{vmatrix} = 124
\end{align}
\begin{align}I_2=
\begin{vmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -5
\end{vmatrix} = -5
\end{align}
We have $det(I_3)>0$ and $det(I_2)\ne0$, $I_2$'s Eigenvalues are $\lambda_1,2 = -1$ and $\lambda_3 = 5$ which indicates that this equation represents an hyperboloid of one sheets
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative vanishes at $x=0$ and $x=2\pi$.
| By Weierstrass Substitution:
By symmetry, we can half the integration interval as
$$
\int_0^{2 \pi} \frac{1}{5-4 \cos x} d x=2 \int_0^\pi \frac{1}{5-4 \cos x} d x
$$
Putting $t=\tan \frac{x}{2}$ transforms the integral into
$$
\begin{aligned}
I & =2 \int_0^\pi \frac{1}{5-\frac{4\left(1-t^2\right)}{1+t^2}} \cdot \frac{2 d t}{1+t^2} \\
& =4 \int_0^{ \infty} \frac{d t}{9 t^2+1} \\
& =\frac{4}{3}\left[\tan ^{-1}(3 t)\right]_0^{\infty} \\
& =\frac{2 \pi}{3}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
| Given equation : $(x^2-3x+1)^2=4x^2-12x+9$ ....(1)
To solve this equation we need the identity,
$a^2-2ab+b^2=(a-b)^2$
on observing the R.H.S. of the equation (1) we can write $4x^2-12x+9$ as
$(2x)^2-2\times2x\times3+3^2$
Now from the above identity, it can be written as $(2x-3)^2$
Thus, equation (1) becomes $(x^2-3x+1)^2=(2x-3)^2$
$\implies$ $(x^2-3x+1)^2-(2x-3)^2=0$
$\implies$ $(x^2-3x+1+2x-3)(x^2-3x+1-2x+3)=0$ $\;\;\;\;$$\because\,a^2-b^2=(a+b)(a-b)$
$\implies$ $(x^2-x-2)(x^2-5x+4)=0$
$\implies$ $(x^2-2x+x-2)(x^2-x-4x+4)=0$
$\implies$ $\{x(x-2)+1(x-2)\}\{x(x-1)-4(x-1)\}=0$
$\implies$ $\{(x-2)(x+1)\}\{(x-1)(x-4)\}=0$
$\implies$ $(x-2)(x+1)(x-1)(x-4)=0$
Thus, x = 2, -1, 1, 4
this is the solution of given equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Simplifying a radical with complex fractions So I understand to simplify this:
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}}
$$
I can just multiply
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}} \cdot\frac{2t^4}{2t^4}
$$
and get
$$
\frac{-3}{t\sqrt{\frac{1}{4t^6} - 1}}
$$
But how do you simplify further getting rid of the complex fraction inside the radical?
| One may write
$$
\begin{align}
\frac{\frac{-3}{2t^4}}{\left|\frac{1}{2t^3}\right|\sqrt{\frac{1}{4t^6} - 1}}&=\frac{\frac{-3}{2t^4}}{\left|\frac{1}{2t^3}\right|\sqrt{\frac{1-4t^6}{4t^6}}}
\\&=\frac{\frac{-3}{2t^4}}{\left|\frac{1}{2t^3}\right|\frac{\sqrt{1-4t^6}}{\sqrt{4t^6}}}
\\&=\frac{\frac{-3}{2t^4}}{\frac{1}{2\left|t\right|^3}\frac{\sqrt{1-4t^6}}{2|t|^3}}
\\&=\frac{\frac{-3}{2t^4}}{\frac{\sqrt{1-4t^6}}{4t^6}}
\\&=\frac{\frac{-3}{2t^4}\times 4t^6}{\sqrt{1-4t^6}}
\\&=-\frac{6t^2}{\sqrt{1-4t^6}}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$ lies on the circle then,
$$6^2+6^2+12g+12f+c=0$$
$$12g+12f+c=-72$$-------(2)
The point $R(2,-2)$ lies on the circle the,
$$2^2+(-2)^2+4g-4f+c=0$$
$$4g-4f+c=0$$-----(3).
Now, please help me from here.
| $\begin{vmatrix}
x^2+y^2&x&y&1\\
5^2+7^2&5&7&1\\
6^2+6^2&6&6&1\\
2^2+(-2)^2&2&-2&1
\end{vmatrix}=0$
$-12(x^2+y^2)+48x+72y+144=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to integrate this integral?. How to integrate $$\int\limits_{1}^{3}\cos(5x^2)\,\mathrm dx \, ?$$ Doesn't seem to work by using intergation by parts or substitution.
| We can simply calculate the indefinite integral:
Substitute $u=\sqrt{\frac{10}{\pi}}x$ to get a Fresnel integral:
$$\int \cos (5x^2)\,\mathrm d x=\sqrt\frac{\pi}{10}\int\cos \left( \frac{\pi}{2}u^2\right)\,\mathrm d u=\sqrt\frac{\pi}{10} C(u)=\sqrt\frac{\pi}{10}\,C\left( \sqrt\frac{10}{\pi} x\right) +C$$
Now use the fundamental theorem of calculus to get:
$$\int_1^3 \cos (5x^2)\,\mathrm d x=\sqrt\frac{\pi}{10}\,C\left( 3\sqrt\frac{10}{\pi}\right)-\sqrt\frac{\pi}{10}\,C\left( \sqrt\frac{10}{\pi} \right)=\\
\boxed{\sqrt\frac{\pi}{10}\left( C\left(3\sqrt\frac{10}{\pi}\right)-C\left(\sqrt\frac{10}{\pi}\right)\right)}\approx 0.124$$
In general, we have:
$$ \int \cos (ax^2)\,\mathrm d x=\sqrt\frac{\pi}{2a}C\left( \sqrt\frac{2a}{\pi}x\right) + C$$
and thus
$$\int_b^c \cos (ax^2)\,\mathrm d x=\sqrt\frac{\pi}{2a}\left( C\left(c\sqrt\frac{2a}{\pi}\right)-C\left(b\sqrt\frac{2a}{\pi}\right)\right)$$
This identity can be derived analogically by substituting $u=\sqrt\frac{2a}{\pi}x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
| Switch the sign of columns 2. and 4., then multiply column 2. and 3. by 3 and finally add them up to get zero in every row:
Col.1 - 3 Col.2 + 3 Col.3 - Col.4 = 0
So columns are linearly dependent, hence the determinant is zero for any $a\ldots d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 5
} |
Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$
I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
| Multiply both the numerator and the denominator by $x(1-x)(1+x)$ and then use the fact that $(1-x)(1+x)=1-x^2$ to see that
\begin{align}
\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}&=\left(\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}\right)\frac{x(1-x)(1+x)}{x(1-x)(1+x)}
\\&=\left(\frac{x^2+(1+x)(1-x)}{(1-x)(1+x)+x^2}\right)\frac{1+x}{1-x}
\\&=\left(\frac{1}{1}\right)\frac{1+x}{1-x}
\\&=\frac{1+x}{1-x}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Show that a $f(r\cos\theta, r\sin\theta)=rh(\theta)$ is continuous on $\mathbb{R}^2$. "Vector Calculus ..." - Hubbard & Hubbard: Exercise 1.20 There's this exercise in Hubbard's book:
Let $ h:\Bbb R \to \Bbb R $ be a $C^1$ function, periodic of period $2\pi$, and define the function $ f:\Bbb R^2 \to \Bbb R $ by
$$f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=rh(\theta)$$
a. Show that $f$ is a continuous real-valued function on $\Bbb R^2$.
b. Show that $f$ is differentiable on $\Bbb R^2 - \{\mathbf 0\}$.
c. Show that all directional derivatives of $f$ exist at $\mathbf 0$ if and only if
$$ h(\theta) = -h(\theta + \pi) \ \text{ for all } \theta $$
d. Show that $f$ is differentiable at $ \mathbf 0 $ if an only if $h(\theta)=a \cos \theta + b \sin \theta$ for some number $a$ and $b$.
I can't find how to prove $ f $ is continuous, I tried to prove
$$ \lim_{\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix} \to \begin{pmatrix}s\cos\phi\\s\sin\phi \end{pmatrix}} f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=s\ h(\phi) $$ for all $s$ and $\phi$.
But I can't do much else.
| First consider the limit of $f$ as you approach some point $\begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}$ for which $s>0$.
\begin{align*}
\lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}}f\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix}&= \lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}} rh(\theta)\\
&=\left(\lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}} r\right)\left( \lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}} h(\theta)\right)\\
&= s\cdot h(\phi) \text{ (by continuity)}\\
&=f\begin{pmatrix}s\cos\phi \\ s \sin \phi\end{pmatrix}.
\end{align*}
Now if you are approaching the origin (which corresponds to $s=0$), $\theta$ need not approach a limit. To show this function is continuous at $(0,0)$, you need to show
$$\lim_{r \to 0}f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=f\begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
regardless of $\theta$. For this you should use the squeeze theorem. Note that since $h$ is continuous and periodic with period $2\pi$, $h$ is bounded on $[0,2\pi]$ and thus on all of $\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nonlinear equations in 3 variables Solve for $x,y\ \mbox{and}\ z$:
$$
\left\{\begin{array}{rcr}
x + y + z & = & 2
\\[1mm]
\left(x + y\right)\left(y + z\right) + \left(y+z\right)\left(z+x\right) +
\left(z + x\right)\left(x + y\right) & = & 1
\\[1mm]
x^{2}\left(y + z\right) + y^{2}\left(x + z\right) + z^{2}\left(x + y\right)
& = & -6
\end{array}\right.
$$
I tried this by adding the second and third equations, and then using the first equation to substitute values. However, this is getting nowhere.
| Way:
$1)$ express LHS-s via elementary symmetric polynomials;
$2)$ consider related cubic equation.
$1)$ Elementary symmetric polynomials for this problem are:
$e_1(x,y,z) = x+y+z$;
$e_2(x,y,z) = xy+xz+yz$;
$e_3(x,y,z) = xyz$.
So,
$$x+y+z= e_1;$$
$$(x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y)\\ =3(xy+yz+xz)+x^2+y^2+z^2 \\
=(x+y+z)^2+(xy+xz+yz) \\ = \left(e_1\right)^2 + e_2;$$
For the last equation: note that
$$(x+y+z)(xy+xz+yz) = x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+3xyz,$$
therefore
$$x^2(y+z) + y^2(x+z) + z^2(x+y) = e_1e_2-3e_3.$$
Now we have:
$$\left\{\begin{array}{l}e_1 = 2;\\(e_1)^2+e_2 =1; \\ e_1e_2-3e_3 = -6;\end{array} \right.
\Rightarrow
\left\{\begin{array}{l}e_1 = 2;\\e_2 =-3; \\ e_3 = 0.\end{array} \right.
$$
$2)$
So (according to Vieta's formulas), triple $(x,y,z)$ is solution of cubic equation
$$w^3-e_1w^2+e_2w-e_3=0.$$
$$w^3-2w^2-3w=0.$$
$$w(w^2-2w-3)=0.$$
Hence, $(x,y,z)=(-1,0,3)$ and all its permutations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Angle of a triangle using trigonometry. If in a triangle ABC , cos A+ cos C= sin B then what is the measurement of angle A? I have tried to solve it using trigonometric identities but failed to solve it.
| $$\cos A + \cos C = \sin(180-A-C)$$
$$\cos A + \cos C = \sin(A+C)$$
$$\cos A + \cos C = \cos A \sin C + \cos C \sin A$$
$$\cos A (1-\sin C) + \cos C (1-\sin A) = 0$$
If $\sin A = 1$ or $\sin C= 1$, we see that this will hold automatically. Otherwise, we have that
$$\frac{\cos A}{1-\sin A} + \frac{\cos C}{1-\sin C} = 0$$
Note that
$$\frac{\cos(90-2x)}{1-\sin(90-2x)} = \frac{\sin(2x)}{1-\cos(2x)} = \frac{2\sin x\cos x}{2\sin^2 x} = \cot x$$
Now, let $A = 90-2\theta,C=90-2\phi$. We then have
$$\cot\theta+\cot\phi =0$$
From a plot of $\cot$, we see that this means
$$\theta = 180n-\phi$$
for some integer $n$.
$$2\theta + 2\phi = 360n$$
$$90-A+90-C = 360n$$
$$A+C = 180-360n$$
Thus, since $A+C$ is a multiple of $180$, this triangle is degenerate.
Because of this, our only cases are when $A$ or $C$ is a right angle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the coefficient of $x^{100}$ in the power series representing the function $\dots$ Find the coefficient of $x^{100}$ in the power series representing the function:
$$f(x)=(x+x^2+x^3+ \cdots) \cdot (x^2+x^3+x^4 \cdots) \cdot (x^3+x^4+x^5 \cdots)$$
Here is what I have so far:
$x+x^2+x^3+ \cdots = x(1+x+x^2+x^3+ \cdots)$ and this equals to the geometric series $\frac{x}{1-x}$
$x^2+x^3+x^4 \cdots = x^2(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^2}{1-x}$
$x^3+x^4+x^5 \cdots = x^3(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^3}{1-x}$
So now we have $F(x)=\left(\frac{x}{1-x}\right)\left(\frac{x^2}{1-x}\right)\left(\frac{x^3}{1-x}\right)$
This is the same as $F(x)=x^6(1-x)^{-1}$
Now we can use the following way to expand and find the coefficient in front of $x^{100}$, $(1+x)^{\alpha}=1+\alpha x + {\alpha \choose 2}x^2 + {\alpha \choose 3}x^3 + \dots$
For our function we get $x^6(1+(-1)(-x) + {-1 \choose 2}(-x)^2 + {-1 \choose 3}(-x)^3 + \dots)$, but $-1 \choose \beta$ where $\beta \ge 2$ will always equal $1$ or $-1$ and I do not think I am doing this correct because according to this the coefficient of $x^{100}$ will be either $1$ or $-1$?
What am I doing wrong?
| As others have noted in the comments, you should have
$$F(x)=\frac{x^6}{(1-x)^3}=x^6\sum_{n\ge 0}\binom{n+2}2x^n=\sum_{n\ge 6}\binom{n-4}2x^n\;.$$
The general rule here is
$$\frac1{(1-x)^{m+1}}=\sum_{n\ge 0}\binom{m+n}mx^n\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When will a parametric solution generate all possible solutions? I was looking for integer solutions to this equation:
$$a^2+b^2+c^2=3d^2$$
And found a parametric solution. Given r, s, t:
$$a=r^2+s^2-t^2+2t(r+s)$$
$$b=r^2-s^2+t^2+2s(t-r)$$
$$c=-r^2+s^2+t^2+2r(t-s)$$
$$d=r^2+s^2+t^2$$
My question is will this generate every possible solution?
| You almost got it. The complete rational parameterization to
$$a^2+b^2+c^2=3\tag1$$
is given by,
$$a=\frac{r^2+s^2-t^2-2t(r+s)}{r^2+s^2+t^2}\\b=\frac{r^2-s^2+t^2-2s(r+t)}{r^2+s^2+t^2}\\c=\frac{-r^2+s^2+t^2-2r(s+t)}{r^2+s^2+t^2}\tag2$$
Proof: For any solution $a,b,c$, one can find $r,s,t$ using the simple formulas,
$$r=\frac{1}{b-1}\\s=\frac{1}{c-1}\\t=\frac{a-1}{(b-1)(c-1)}\tag3$$
Example: Using Jagy's example, we have $a,b,c = \frac{1}{7}, \frac{5}{7},
\frac{11}{7}$. Plugging this into $(3)$, we get $r,s,t = \frac{-7}{2}, \frac{7}{4}, \frac{21}{4}$. Plugging this into $(2)$, we recover the $a,b,c$.
Note: Division by zero is easily avoided by using $(\pm1)^2+(\pm1)^2+(\pm1)^2=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction
$$ 21 \mid (2^{4n}+5) $$
So I assume:
$ 2^{4k}+5= 21p$
to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ =
$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 16 \cdot 21p - 80+5 = 16 \cdot 21p - 75 $
But its not divisible by 21. Whats I doing wrong ?
| The statement is true for $n=1$. Suppose it holds for $k$: $2^{4k}+5=21p$, so $2^{4k}=21p-5$, so
$$
2^{4(k+1)}+5=2^{4k}\cdot 16+5=
16(21p-5)+5=21\cdot 16p-80+5=21\cdot16p-75
$$
Ops! Something seems wrong. We have proved that
if $21\mid 2^{4k}+5$, then $21\nmid 2^{4(k+1)}+5$
Actually, $3\mid 2^{4n}+5$ (prove it), but in general $7\nmid 2^{4n}+5$; indeed,
$$
2^{4n}+5=16^n+5\equiv 2^n+5\equiv2^n-2\equiv2(2^{n-1}-1)\pmod{7}
$$
Now
\begin{align}
2^0-1&\equiv 0\pmod{7}\\
2^1-1&\equiv 1\pmod{7}\\
2^2-1&\equiv 3\pmod{7}\\
2^3-1&\equiv 0\pmod{7}
\end{align}
so $2^{4n}+5$ is divisible by $7$ if and only if $3\mid(n-1)$, therefore when $n=3m+1$, for some $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simple way to calculate $n! \pmod p$ I have the exercise "Calculate $10! \pmod{13}$".
I have the following two approaches to solve the exercise:
*
*Brute force approach
$$
1! \equiv 1 \pmod{13} \\
2! = 2 \cdot 1! \equiv 2 \cdot 1 \equiv 2 \pmod{13} \\
3! = 3 \cdot 2! \equiv 3 \cdot 2 \equiv 6 \pmod{13} \\
\cdots \\
10! = 10 \cdot 9! \equiv 10 \cdot 11 = 110 = 8 \cdot 13 + 6 \equiv 6 \pmod{13}
$$
*Approach using Wilson's theorem:
Wilson's theorem states that
$$p \in \mathbb{P} \implies (p-1)! \equiv -1 \pmod p$$
For my exercise:
$$13 \in \mathbb{P} \implies \\
(13-1)! = 12! = 10!\cdot 11 \cdot 12 \equiv -1 \pmod{13} \implies \\
10! \equiv -(11 \cdot 12)^{-1} \pmod{13}
$$
Using Fermat's little theorem
$$
a^p \equiv a \pmod p \implies a^{p-2} \cdot a \cdot a \equiv a^{-1} \cdot a \cdot a \pmod p \implies a^{p-2} \equiv a^{-1} \pmod p \\
$$
For my exercise:
$$10! = -(11 \cdot 12)^{-1} \equiv \\
-(11 \cdot 12)^{13-2} = -(11 \cdot 12)^{11} \equiv \\
-(-2 \cdot -1)^{11} = -2^{11} = \\
-2^6 \cdot 2^5 \equiv 1 \cdot 2^5 = \\
32 \equiv 6 \pmod{13} \\
$$
Both approaches look quite bulky.
In the first method I have to make $O(n)$ multiplications.
In the second method I have to make $O(p-n)$ multiplications which is smaller than in the first method, but also can be huge number for big $p$ and $n$.
Is there a way to improve my solution?
Is there an efficient way to calculate $n! \pmod p$ for big $n$ and $p$?
| Semi brute force. $14\equiv 1 \mod 13$ so toss out $2,7$. $27 \equiv 1$ so toss out $3,9$. $40 \equiv 1$ so toss out $4,10$ and $8,5$. What do we have left.
Just $6$. $10! \equiv 6\mod 13$
because $10! =(2*7)(3*9)(4*10)6 (5*8)\equiv 6 \mod 13$
Hmm, that's not very satisfying, is it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Show $\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$ I want to verify the following inequality:
Let $a, b$ be non negative number
$$\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$$
I decided to analyse the sign of $$\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 - 2\left(\frac{a}{b} + \frac{b}{a}\right)$$
But I'm not getting anywhere.
I'm having trouble deciding whether $\dfrac{a^2(a-2)+b^2(b-2)}{ab} +3$ is positive or negative.
| Let $t=a/b+b/a$. Then this becomes $t^2+1\geq 2t$, which is same as $(t-1)^2 \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
| As is clear, this question is invariant under horizontal displacement i.e. you can replace $f(x)=(x-\dfrac{a}{2})^2+1-\dfrac{a^2}{4}$ with $f(x)=x^2+1-\dfrac{a^2}{4}$. Here after we use the latter notation. Define $b=1-\dfrac{a^2}{4}$therefore $$f(x)=x^2+b$$assume such a $x_0$ exists. We cannot have both $x_0$ and $x_0+2$ are positive or both negative. To show that if they are both positive we have $$-0.25<x_0^2+b<0.25\qquad(1)\\-0.25<(x_0+2)^2+b<0.25\qquad(2)$$but this is a contradiction since $$(x_0+2)^2+b=x_0^2+b+4x_0+4>-0.25+4x_0+4>3.75>0.25$$to the same argument they can't both be negative so we have $$x_0<0\\x_0+2>0$$the function is strictly decreasing when $x<0$ and strictly increasing when $x>0$ therefore the inequalities $(1)$ and $(2)$ impose that $$-\sqrt{0.25-b}\le x_0\le -\sqrt{-0.25-b}\\\sqrt{-0.25-b}\le x_0+2\le\sqrt{0.25-b}$$or equivalently $$-\sqrt{0.25-b}\le x_0\le -\sqrt{-0.25-b}\\\sqrt{-0.25-b}-2\le x_0\le \sqrt{0.25-b}-2$$both intervals are of equal length therefore such a $x_0$ exists and satisfies both inequalities iff the intervals intersect i.e.$$\text{either }\quad -\sqrt{0.25-b}\le \sqrt{-0.25-b}-2\le -\sqrt{-0.25-b}\\\text{or }\quad-\sqrt{0.25-b}\le \sqrt{0.25-b}-2\le -\sqrt{-0.25-b}$$which leaves us with $$-1.25\le b\le-1-\dfrac{1}{64}\\\text{or}\\-1-\dfrac{1}{64}\le b\le -0.75$$$$-1.25\le b \le -0.75$$which means that $$7\le a^2\le 9$$which means that $$\Large\sqrt 7\le |a|\le3$$You can see a sketch of the answer for $b=-1.25$ and $b=-0.75$ below
where $x_0$ and $x_0+2$ are at the ends of the blue line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 6
} |
Precalc coursework - Is this an error? I am enrolled in a precalc math course at a university studying online. I haven't studied math in over 35 years so I'm a bit rusty. Why would my online math coursework have errors? I end up torn between feeling like I must just not understand something and certainty that there is an error in my professor's notes. I turn to you for confirmation and hopefully confidence.
I attach an image of the professor's solution, please could you confirm for me whether my belief that she is missing an X factor in her work and she then continues solving the equation incorrectly. Who is right?
| You are right. There is a missing $x.$ The final result should be $\dfrac{x+4}{2x^2}.$ Note that
$$\begin{align}\frac{1}{x+3}\left(\frac{x+7}{2x}+\frac{6}{x^2}\right)& = \frac{1}{x+3}\left(\frac{(x+7)\cdot x}{2x\cdot x}+\frac{6\cdot 2}{x^2\cdot 2}\right)\\ &= \frac{1}{x+3}\left(\frac{x^2+7 x}{2x^2}+\frac{12}{2x^2}\right) \\ &= \frac{1}{x+3}\frac{x^2+7 x+12}{2x^2} \\ &= \frac{x^2+7 x+12}{2x^2(x+3)} \\ &= \frac{(x+4)(x+3)}{2x^2(x+3)}\\ &= \frac{x+4}{2x^2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$
I think $uvw$ does not help here.
My another similar inequality is very easy:
with the same condition prove that:
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$
My proof:
By AM-GM $(a+b+c)^3=a^3+b^3+c^3+24\geq9\sqrt[9]{(a^3+b^3+c^3)\cdot3^8}$
and we are done!
Thank you!
| Another solution using KKT conditions (or Lagrange multiplier)
Let
\begin{align}
L = \ln \frac{a+b+c}{3} - \frac{1}{53}\ln \frac{a^4+b^4+c^4}{3}
- t ((a+b)(b+c)(c+a) - 8).
\end{align}
From $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial b} = \frac{\partial L}{\partial c} = 0$, we have
\begin{align}
0 = \frac{\partial L}{\partial a} - \frac{\partial L}{\partial b}=
\frac{\left(a - b\right)\, \left((53a^5+53a^4b+53ab^4+53ac^4+53b^5+53bc^4)t-4a^2-4ab-4b^2\right)}{53(a^4 + b^4 + c^4)},\quad \cdots\cdots\cdots (2)\\
0 = \frac{\partial L}{\partial b} - \frac{\partial L}{\partial c}
= \frac{\left(b - c\right)\, \left((53a^4b+53a^4c+53b^5+53b^4c+53bc^4+53c^5)t-4b^2-4bc-4c^2\right)}{53(a^4 + b^4 + c^4)}. \quad \cdots\cdots\cdots(3)
\end{align}
We claim that two of $a, b, c$ are equal. Otherwise, from (2), we have
$$t= \frac{4(a^2+ab+b^2)}{53(a^5 + a^4\, b + a\, b^4 + a\, c^4 + b^5 + b\, c^4)}$$
which, when inserted into (3), results in
$$\frac{4}{53}\frac{\left(a - c\right)\, \left(b-c\right)\, \left(a\, b + a\, c + b\, c\right)}{\left(a + b\right)\, \left(a^4 + b^4 + c^4\right)} =0.$$
Contradiction.
Further we claim that $a = b = c$. Otherwise, WLOG, assume that $a = b \ne c$, from (3), we have
$$t = \frac{4}{53}\frac{a^2 + a\, c + c^2}{2\, a^5 + 2\, a^4\, c + a\, c^4 + c^5}$$
which results in
$$0 = \frac{\partial L}{\partial a} = -\frac{1}{53}\frac{ - 74\, a^4 + 48\, a^3\, c + 48\, a^2\, c^2 + 24\, a\, c^3 - 49\, c^4}{\left(2\, a^4 + c^4\right)\, \left(2\, a + c\right)}.$$
However, $- 74\, a^4 + 48\, a^3\, c + 48\, a^2\, c^2 + 24\, a\, c^3 - 49\, c^4\ne 0$
which follows from $74x^4-48x^3-48x^2-24x+49 > 0$ for all real numbers $x$. Contradiction. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Algebraic expansion of $(1+x)^{1/x}$ is? i have to solve
$ \lim_{x\to 0} \frac{{(1+x})^{1/x}-e+ex/2}{x^2}$ and in the hint it is advised to use the algebraic expansion of $(1+x)^{1/x}$.
Also i know that the algebraic expansion of $(1+x)^{m} = 1+mx+m(m-1){x^2}/2+...$ but this is in the case where m is an integer.
So please help me here with the limit.
| Note that you can develop
$$
\begin{gathered}
\left( {1 + x} \right)^{\,1/x} = \exp \left( {\frac{1}
{x}\ln \left( {1 + x} \right)} \right) = \exp \left( {1 - \frac{x}
{2} + \frac{{x^{\,2} }}
{3} + O\left( {x^{\,3} } \right)} \right) = \hfill \\
= \exp \left( 1 \right)\exp \left( { - \frac{x}
{2}} \right)\exp \left( {\frac{{x^{\,2} }}
{3}} \right)\exp \left( {O\left( {x^{\,3} } \right)} \right) = \hfill \\
= e\left( {1 - \frac{x}
{2} + \frac{{x^{\,2} }}
{8}} \right)\left( {1 + \frac{{x^{\,2} }}
{3}} \right) + O\left( {x^{\,3} } \right) = e\left( {1 - \frac{x}
{2} + \frac{{11\;x^{\,2} }}
{{24}} + O\left( {x^{\,3} } \right)} \right) \hfill \\
\end{gathered}
$$
from which
$$
\begin{gathered}
\frac{1}
{{x^{\,2} }}\left( {\left( {1 + x} \right)^{\,1/x} - \left( {1 - x/2} \right)e} \right) = \hfill \\
= \frac{1}
{{x^{\,2} }}\left( {e\left( {1 - \frac{x}
{2} + \frac{{11\;x^{\,2} }}
{{24}}} \right) - \left( {1 - x/2} \right)e + O\left( {x^{\,3} } \right)} \right) = \hfill \\
= \frac{{11}}
{{24}}e + O\left( x \right) \hfill \\
\end{gathered}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Integration of $\frac{2x^2}{\sqrt{x^2-1}}\, dx$ Integration of $\dfrac{2x^2}{\sqrt{x^2-1}}\,\mathrm dx$
My try:
Let $u=\sqrt{x^2-1}$
Then $x=\sqrt{u^2+1}$
$x \,\mathrm dx =u\,\mathrm du$
$$\int \frac{2u(u^2+1)}{u\sqrt{u^2+1}} \, \mathrm du$$
$$=2\int \sqrt {u^2+1} \, \mathrm du$$
True ? and what about the last integration?
| my hint:
$$\int \frac{2x^2}{\sqrt{x^2-1}}\ dx=\int \frac{2(x^2-1)+2}{\sqrt{x^2-1}}\ dx=2\int\sqrt{x^2-1}\ dx+2\int\frac{1}{\sqrt{x^2-1}}\ dx$$
both integrals can be evaluated by substituting $u=\sec \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Arithmetic Mean/Geometrix Mean Inequality of Degree 3 For $a,b$ and $c\geq0$;
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
Is there a simple way to deduce this result from the degree 2 verson?
That is:
$$\frac{a+b}{2}\geq\sqrt{ab}$$
Some sort of substitution involving $a,b,c \ $ for one of the variables was what I had in mind, but I can't seem to figure it out!
| I think, the previous method is the best (with using AM-GM for two variables),
but we can make also the following proof by the same AM-GM.
Since for non-negatives $x$, $y$ and $z$ by AM-GM we have
$\frac{x^2+y^2}{2}\geq xy$ and $\frac{x^2+z^2}{2}\geq xz$ and $\frac{y^2+z^2}{2}\geq yz$,
after summing we obtain $x^2+y^2+z^2-xy-xz-yz\geq0$.
Now let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, $\frac{a+b+c}{3}-\sqrt[3]{abc}=\frac{x^3+y^3+z^3-3xyz}{3}=\frac{(x+y+z)(x^2+y^2+z^2-xy-xz-yz)}{3}\geq0$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
The matrix $e^A$ is defined by $e^A=\Sigma_{k=0}^{\infty}\frac {A^k}{k!}$ Suppose M=$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. Calculate $e^M$ The matrix $e^A$ is defined by $e^A=\Sigma_{k=0}^{\infty}\frac {A^k}{k!}$ Suppose M=$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. Calculate $e^M$.
I did some calculating with real values and I got the iteration of values: $\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, $\begin{bmatrix}2 & 1\\0 & 2\end{bmatrix}$, $\begin{bmatrix}\frac 5 2 & 2\\0 & \frac 5 2 \end{bmatrix}$, etc.
I believe that the solution is $\begin{bmatrix}e & e\\0 & e\end{bmatrix}$, but can't prove it.
| Observe
\begin{align}
M=
\begin{pmatrix}
1& 0\\
0 & 1
\end{pmatrix}
+
\begin{pmatrix}
0& 1\\
0 & 0
\end{pmatrix}=: I+ X
\end{align}
where $[X, I] = 0$ (of course since identity commutes with anything). Hence it follows
\begin{align}
\exp[M] = \exp[I]\exp[X].
\end{align}
Next, observe
\begin{align}
X^2 = 0 \ \ \Rightarrow \ \ \exp[X] = I + X+ \frac{1}{2!}X^2+\ldots = I+X
\end{align}
which means
\begin{align}
\exp[M] = eI \ast(I+X).
\end{align}
Thus, we have the desired solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Summation Proof prove that:
I am having trouble coming up with a rigorous proof for this.Can I use induction? Can anyone demonstrate?
$$\sum_{i={2^n+1}}^{{2^{n+1}}}{1/i}\ge 1/2$$
| $$
\sum_{i = 2^n + 1}^{2^{n+1}} \frac{1}{i} = \frac{1}{2^n + 1} + \frac{1}{2^n + } + \cdots + \frac{1}{2^n + 2^n}
$$
$$
> \frac{1}{2^n + 2^n} + \frac{1}{2^n + 2^n} + \cdots + \frac{1}{2^n + 2^n} = \frac{2^n}{2^n + 2^n} = \frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit involving nested roots How do I solve it? Is there some crafty algebraic step I need to follow?
I'm guessing it's -∞ in the end..
$$ \lim_{x\to-∞} \sqrt{x^2-\sqrt{x^2+1}}+x $$
| Rewrite it as follows: for $x< 0$,
$$\begin{align}
\sqrt{x^2-\sqrt{x^2+1}}+x &= \sqrt{x^2-|x|\sqrt{1+\frac{1}{x^2}}}+x = |x|\sqrt{1-\frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}}+x \\
&= -x\left(1-\sqrt{1-\frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}}\right)
\end{align}$$
Writing $u(x)\stackrel{\rm def}{=} \frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}$, we have $u(x)\xrightarrow[x\to-\infty]{} 0$, and by a first-order Taylor expansion
$$
\sqrt{1-u(x)} = 1-\frac{u(x)}{2} + o(u(x))
$$
when $x\to-\infty$. Therefore, we get, when $x\to-\infty$,
$$\begin{align}
\sqrt{x^2-\sqrt{x^2+1}}+x
&= x\left(1-\sqrt{1-u(x)}\right)
= x\left( 1-1+\frac{u(x)}{2} + o(u(x)) \right) \\
&= \frac{1}{2}xu(x) + o(xu(x))
\end{align}$$
and since $xu(x)=-\sqrt{1+\frac{1}{x^2}} + o(1) \xrightarrow[x\to-\infty]{} -1$, we eventually obtain
$$
\sqrt{x^2-\sqrt{x^2+1}}+x
\xrightarrow[x\to-\infty]{} \frac{1}{2}\cdot -1 = -\frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $ Let $x,y,z > 0$ and $xyz=8.$ Prove that
$$
\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12
$$
I have tried with AM–GM inequality but no result.
| We'll prove that $\sum\limits_{cyc}\frac{(x+y-1)^2}{z}\geq\frac{27}{2}$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $w=2$ and we need to prove that $f(v^2)\geq0$, where
$f(v^2)=9u^2v^2-5uw^3-3uv^2w^3-\frac{5}{4}w^4+\frac{1}{4}v^2w^2$.
But $f$ is a linear function, which says that it's enough to prove our inequality
for an extremal value of $v^2$, which happens for equality case of two variables.
Indeed, $x$, $y$ and $z$ are positive roots of the equation $(X-x)(X-y)(X-z)=0$ or
$X^3-3uX^2+3v^2X-w^3=0$ or $3v^2X=-X^3+3uX^2+w^3$.
Thus, the last equation has three positive roots, which says that a line $Y=3v^2X$
and a graph of $Y=-X^3+3uX^2+w^3$ have three common points (draw it!),
which says that $v^2$ gets an extremal value, when a graph of $Y=3v^2X$
is a tangent line to the graph of $Y=-X^3+3uX^2+w^3$,
which happens for equality case of two variables.
Let $y=x$ and $z=\frac{8}{x^2}$.
We obtain $(x-2)^2(4x^7+12x^6+33x^5+100x^4+128x^3+128x^2+256x+256)\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating the improper integral $\int_0^\infty \frac{x\cos x-\sin x}{x^3} \cos(\frac{x}{2}) \mathrm dx $ I've been working through the following integral and am stumped:
$$\int_0^\infty \frac{x\cos x-\sin x}{x^3}\cos\left(\frac{x}{2}\right)\mathrm dx$$
Given the questions in my class that have proceeded and followed this integral, I believe that this is some form of Fourier transform/integral. However, it doesn't look like any of the content surrounding it. That is, there is no $e^{-ikx}$ or $g(k)$ or anything else that I'm familiar with.
I know that it's an even function, but that's about as far as I can get. If I try to split it over the subtraction, I get two non-converging integrals, so that wasn't much help either. I've been throwing lots of trig identities at it but nothing familiar has appeared yet.
Any help would be greatly appreciated.
Thank you.
| Inspired by Felix Marin's calculation using integration by parts.
Observe
\begin{align}
\int^\infty_0 \frac{x\cos x-\sin x}{x^3}\cos\frac{x}{2}\ dx=&\ \frac{1}{2}\int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^3}e^{ix/2}\ dx\\
=&\ \frac{-1}{4} \int^\infty_{-\infty} [x\cos x-\sin x] e^{ix/2}\ d\left(\frac{1}{x^2} \right).
\end{align}
Using integration by parts, we have
\begin{align}
\frac{-1}{4} \int^\infty_{-\infty} [x\cos x-\sin x] e^{ix/2}\ d\left(\frac{1}{x^2} \right)=&\ \frac{1}{4} \int^\infty_{-\infty}d([x\cos x-\sin x]e^{ix/2}) \frac{1}{x^2}\\
=&\ \frac{-1}{4} \int^\infty_{-\infty}\frac{\sin x}{x}e^{ix/2}\ dx + \frac{i}{8} \int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^2}e^{ix/2}\ dx.
\end{align}
Now, observe
\begin{align}
\int^\infty_{-\infty}\frac{\sin x}{x}e^{ix/2}\ dx = \mathcal{F}^{-1}\left[\operatorname{sinc\left(\frac{x}{\pi}\right)}\right]\left(\frac{1}{2}\right) = \pi \mathcal{F}^{-1}[\operatorname{sinc}]\left( \frac{1}{2\pi}\right) = \pi.
\end{align}
Next, observe
\begin{align}
\int^\infty_{-\infty} \frac{x\cos x-\sin x}{x^2} e^{ix/2}\ dx =&\ \int^\infty_{-\infty} \frac{d}{dx}\left( \frac{\sin x}{x}\right) e^{ix/2}\ dx\\
=&\ -\frac{i}{2}\int^\infty_{-\infty}\frac{\sin x}{x} e^{ix/2}\ dx = -\frac{i\pi}{2}.
\end{align}
Hence combining everything yields
\begin{align}
\int^\infty_0 \frac{x\cos x-\sin x}{x^3} \cos \frac{x}{2}\ dx = -\frac{\pi}{4} + \frac{\pi}{16} = -\frac{3\pi}{16}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
} |
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking $$x=\sqrt{2}y$$ then using integration by parts, we have
$$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$
But how to calculate this integral? I would appreciate your help
| An elementary approach
\begin{align}
I &= \int_0^1 \frac{\arcsin x\arcsin\frac x{\sqrt2}}{\sqrt{2-x^2}} \, dx \\
&= \frac12 \int_0^1 \arcsin x \ d\left(\arcsin^2\frac x{\sqrt2} -\frac{\pi^2}{16}\right)\\
&\overset{ibp}= \frac12 \int_0^1 \left(\frac{\pi^2}{16}-\arcsin^2\frac x{\sqrt2} \right)\frac1{\sqrt{1-x^2}}dx
\end{align}
Substitute $x=\sqrt2\sin\left(\frac\pi4-\theta\right)$ and then $\theta\to\frac\pi2-\theta$
\begin{align}
I &= \frac1{4\sqrt2}\int_0^{\pi/2}\theta\left(\frac\pi2-\theta\right)\left(\sqrt{\tan\theta}+\sqrt{\cot\theta}\right)
\overset{\tan \theta =t^2}{d\theta} \\
&= \frac1{2\sqrt2} \int_0^\infty
\frac{(t^2+1)\tan^{-1} t^2 \cot^{-1} t^2}{t^4+1}dt\\
&= \frac1{2\sqrt2}\left( \int_0^\infty
\frac{\tan^{-1} t^2 \cot^{-1} t^2}{t^2+\sqrt2t+1}dt
+ \sqrt2\int_0^\infty\frac{t\tan^{-1} t^2 \cot^{-1} t^2}{t^4+1} dt\right)\\
&= \frac1{2\sqrt2}\left(\frac\pi{4\sqrt2}\ln^22+\sqrt2\cdot \frac{\pi^3}{48} \right)=\frac\pi{16}\left( \ln^22+\frac{\pi^2}{12}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find all k such that $\sigma (k)=165$ where $\sigma$ is the sum of divisors $165=(3) (5) (11)$
$\sigma (p^a)$=$p^{a+1}-1 \over {p-1}$=$3$
$p^{a+1}-1$=$3p-3$
$p^{a+1}=3p-2$
Got stuck here. How do I proceed?
| Since it seems that you are looking for the "hard" way:
Let $n$ be a positive integer such that $\sigma(n)=165.$ Then $\prod_{p\mid n}(p^{\alpha_p+1}-1)/(p-1)=3\cdot5\cdot11,$ where the product is taken over all primes dividing $n$ and where $\alpha_p$ is the exponent of $p$ in the prime decomposition of $n.$ Now note that if $p\mid n$ then $\alpha_p+1\geqslant2$ so $(p^{\alpha_p+1}-1)/(p-1)>1.$ Hence, since the RHS of the equality above has exactly three prime factors, $n$ has at most three prime factors. If $n=p^a$ for some prime $p$ and some integer $a>0$ then $\sigma(n)=(p^{a+1}-1)/(p-1)$ so $164=2^2\cdot41=p(165-p^a)$ so $p\in\{2,41\}.$ It is easy to check that none of these works. Thus $n$ has either two or three prime factors. If $n=p^aq^b,$ with $p$ and $q$ primes, $p\neq q$ and $a,b>0$ then we have three possibilities. If we set $\alpha:=(p^{a+1}-1)/(p-1)$ and $\beta:=(q^{b+1}-1)/(q-1),$ the three possibilities are:
$(i)$ $\alpha=15$ and $\beta=11.$ Then $14=2\cdot7=p(15-p^a)$ so $p\in\{2,7\}.$ In the same way we find that $q\in\{2,5\}.$ If $p=2$ then $a=3$ and also since $p\neq q$ and $10=2\cdot5=q(11-q^b)$ then $q=5$ so $5^b=9,$ which is impossible. If $p=7$ then $7^a=13,$ which is impossible.
$(ii)$ $\alpha=33$ and $\beta=5.$ Then $32=2^5=p(33-p^a)$ so $p=2.$ Also $4=2^2=q(5-q^b)$ so $q=2,$ which contradicts the fact that $p\neq q.$
$(iii)$ $\alpha=3$ and $\beta=55.$ Then $2=p(3-p^a)$ so $p=2$ and $a=1.$ Also $54=2\cdot3^3=q(55-q^b)$ and since $q\neq p$ then $q=3$ and thus $3^b=37,$ which is impossible.
It remains to see what happens when $n=p^aq^br^c,$ where $p,q,r$ are pairwise distinct primes and $a,b,c$ are positive integers. Now note that there is only one possibility, which is $(p^{a+1}-1)/(p-1)=3,$ $(q^{b+1}-1)/(q-1)=5$ and $(r^{c+1}-1)/(r-1)=11.$ Then $p=q=2,$ which contradicts the fact that $p\neq q. $ Thus there is no $n$ with $\sigma(n)=165.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to solve a particular well ordering problem I want to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$ with well ordering principle.
I know that $$n \in N $$
because the numbers 1,2,3... are discrete and are positive. I was thinking of using a set of natural numbers that don't satisfy the equation and proving by contradiction but I'm not sure that works.
| Yes, you can make it like that. It is not different than induction, though.
Let $E$ be the set of $n\in\mathbb N$ such that your equality does not hold. Note that $1\not\in E$. Let $k\geq2$ be the minimum of $E$ (here is where one uses the principle). Then $k-1\not\in E$, so it satisfies the formula:
$$
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{(k-1)k}=\frac{k-1}{k}.
$$
But then
\begin{align}
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{k(k+1)}&=
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{(k-1)k}+ \frac{1}{k(k+1)}\\ \ \\
&=\frac{k-1}k+\frac{1}{k(k+1)}=\frac{(k-1)(k+1)+1}{k(k+1)}\\ \ \\
&=\frac{k^2}{k(k+1)}=\frac{k}{k+1}.
\end{align}
Then $k$ would satisfy the equality a contradiction. That is, the minimum of $E$ cannot exist. So it has to be that $E$ is empty, and the formula holds for all $n\in\mathbb N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
| Yes, $x+1$ is a factor of $x^{2016}-1$ and
$$x^{2016}-1=(x+1)(x^{2015}-x^{2014}+x^{2013}-\cdots-x^2+x-1)$$
Then
$$G(x)=1+x^2+x^4+\dots+x^{2014}$$
More generally, $x-r$ is a factor of $P(x)$ if and only if $P(r)=0$. In this case, $P(x)=x^{2016}-1$ and $P(-1)=0$, so $x+1$ is a factor of $x^{2016}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the average rate of change of $S(x) = -2x^2 + 14x - 12$ I have to show that the average rate of change of $S(x) = -2x^2 + 14x - 12$ in the interval $[x,x+h]$ is $-4x - 2h + 14$ and so far I did:
$$A(x) = \frac{S(x+h)-S(x)}{(x+h)-(x)} = \frac{(-2(x+h)^2+14(x+h)-12)-(-2x^2+14x-12)}{(x+h)-x} = \frac{-2(x+h)^2+14(x+h)-12+2x^2-14x+12}{-x+x+h} = \frac{-2(x+h)^2+14(x+h)+2x^2-14x}{h} = \frac{-2h^2+14h}{h} = \frac{-2h^2+14h}{h}\\ = -2^2 +14 = 4+ 14 = 18$$
I know this is totally wrong. What did I do wrong? How do I solve this?
| Starting from
$$\frac{-2(x+h)^2+14(x+h)+2x^2-14x}h$$
expand all terms in the numerator and simplify:
$$=\frac{-2(x^2+2hx+h^2)+14x+14h+2x^2-14x}h$$
$$=\frac{-2x^2-4hx-2h^2+14h+2x^2}h$$
$$=\frac{-4hx-2h^2+14h}h$$
$$=-4x-2h+14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a convergence for the series $ \sum_{i=0}^{\infty} \frac{(x-i)^i}{i!} $ The following series converges to exponential.
$\sum_{i=0}^{\infty} \frac{x^i}{i!} = e^{x}$
Is the convergence of the following series known?
$\sum_{i=0}^{\infty} \frac{(x-i)^i}{i!}$
| $$a_n=\frac{(x-n)^n}{n!}\implies A=\frac{a_{n+1}}{a_n}=\frac1{n+1 }\frac{ (x-n-1)^{n+1}} {(x-n)^n }=\frac{-1}{n+1 }\frac{ (n+1-x)^{n+1}} {(n-x)^n }$$ $$B=\frac{ (n+1-x)^{n+1}} {(n-x)^n }\implies\log(B)=(n+1)\log(n+1-x)-n\log(n-x)$$ Now, for large values of $n$, use Taylor$$p \log (p-x)=p \log \left(p\right)-x-\frac{x^2}{2 p}-\frac{x^3}{3
p^2}+O\left(\frac{1}{p^3}\right)$$ Apply to each term and simplify to get $$\log(B)=1+\log \left(n\right)+\frac{1}{2
n}+\frac{\frac{x^2}{2}-\frac{1}{6}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$B=e^{\log(B)}=e n+\frac{e}{2}+\frac{e \left(12 x^2-1\right)}{24
n}+O\left(\frac{1}{n^2}\right)$$ $$A=\frac{-B}{n+1}=-e+\frac{e}{2 n}-\frac{e \left(12 x^2+11\right)}{24
n^2}+O\left(\frac{1}{n ^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$
$$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$
$$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$
Now how can i solve it after that , Help required, Thanks
| Using $$\sin (a+b+c)-\sin a-\sin b-\sin c $$
$$= 2\cos\left(\frac{2a+b+c}{2}\right)\sin \left(\frac{b+c}{2}\right)-2\sin \left(\frac{b+c}{2}\right)\cos \left(\frac{b-c}{2}\right)$$
So $$ = 2\sin \left(\frac{b+c}{2}\right)\left[\cos \left( \frac{2a+b+c}{2}\right)-\cos \left(\frac{b-c}{2}\right)\right]$$
$$ = -4\sin \left(\frac{a+b}{2}\right)\sin \left(\frac{b+c}{2}\right)\sin \left(\frac{a+c}{2}\right)<0,$$
Bcz given $\displaystyle a,b,c \in \left(0,\frac{\pi}{2}\right)$. So we get $\displaystyle \frac{a+b}{2},\frac{b+c}{2}\;,\frac{c+a}{2}\in \left(0,\frac{\pi}{2}\right)$
So we get $$\sin (a+b+c)<\sin a+\sin b+\sin c\Rightarrow \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving Recurrence Relation T(n) = T(n-1) + n + 2 I am getting stuck solving this recurrence relation.
T(1) = 1
T(n) = T(n-1) + n + 2
I continued to expand it:
$T(n) = T(n-1) + n + 2 = [T(n-2) + (n-1) + n + 2] + 2$
$= T(n-2) + (n-1) + n +4 = [T(n-3) + (n-2) + (n-1) + n + 2] + 4$
$= T(n-3) + (n-2) + (n-1) + n + 6$
$= T(n-(n-1)) + (\sum_{i=0}^{n-2} n-i) + 2(n-1)$
$= 1 + (\sum_{i=0}^{n-2} n-i) + 2n-2$
This is where I get stuck. Did I do a step wrong?
| HINT:
$$T(n) - T(n-1) = n +2 = T(n-1) - T(n-2) + 1 \implies T(n) - 2T(n-1) + T(n-2) = 1$$
Now in the same manner:
$$T(n) - 2T(n-1) + T(n-2) = 1 = T(n-1) - 2T(n-2) + T(n-3) \implies $$
$$T(n) - 3T(n-2) + 3T(n-2) - T(n-3) = 0$$
Now this is a homogenous recursive relation, which can be easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Write the given Determinant as the product of two Determinants
Question Statement:-
Prove that:-
$$\begin{vmatrix}
b+c-a-d & bc-ad & bc(a+d)-ad(b+c)\\
c+a-b-d & ca-bd & ca(b+d)-bd(c+a)\\
a+b-c-d & ab-cd & ab(c+d)-cd(a+b)\\
\end{vmatrix}\qquad\qquad=2(a-b)(b-c)(c-a)(a-d)(b-d)(c-d)$$
Attempt at a solution:-
I tried to write the given determinant as the product of two other determinants but was not able to come up with anything worthwhile. My attempt was to think on seeing the first column that the determinant would look somehow like this
$$\begin{vmatrix}
b+c & -1 & ?\\
c+a & -1 & ?\\
a+b & -1 & ?\\
\end{vmatrix}\begin{vmatrix}
1 & a+d & ?\\
1 & b+d & ?\\
1 & c+d & ?\\
\end{vmatrix}$$
The question marks represent those entries of the determinant which I was not able to come up with to obtain the original matrix.
It will be very helpful if you could guide me in the right direction to obtain the the determinants whose product gives the original determinant.
If you have an easier method then it would be very helpful too.
| I don't know how to do it your way, but here is another one. It goes by row-reduction.
In a first step, we perform row reduction in the following way: replace $R_1$ with $R_1-R_2$ and $R_3$ with $R_3-R_1$. And we want to replace $R_2$ with $R_1+R_2$. This can be obtained as $$2\left(R_2+\frac{R_1-R_2}2\right)=R_1+R_2.$$ The net effect is that after doing $R_2+(R_1-R_2)/2$ we need to multiply by $2$. This requires us to also multiply by $1/2$. Then our determinant $D$ is equal to
$$
D=\frac12\,\begin{vmatrix}
2b-2a& bc+bd-ad-ac&2bcd-2acd\\
2c-2d& ac+bc-ad-bd& 2abc-2abd\\
2b-2c&ab+bd-ac-cd&2abd-2acd
\end{vmatrix}.
$$
Factoring in each entry, we get
$$
D=\frac12\,\begin{vmatrix}
2(b-a)& (b-a)(c+d)&2cd(b-a)\\
2(c-d)& (a+b)(c-d)& 2ab(c-d)\\
2(b-c)&(a+d)(b-c)&2ad(b-c)
\end{vmatrix}.
$$
Extracting the common factor in each row,
$$
D=\frac12\,(b-a)(c-d)(b-c)\begin{vmatrix}
2& c+d&2cd\\
2& a+b& 2ab\\
2&a+d&2ad
\end{vmatrix}.
$$
Now replace $R_2$ with $R_2-R_3$ and $R_3$ with $R_3-R_1$:
$$
D=\frac12\,(b-a)(c-d)(b-c)\begin{vmatrix}
2& c+d&2cd\\
0& b-d& 2a(b-d)\\
0&a-c&2d(a-c)
\end{vmatrix}.
$$
Now we can calculate directly. But if we first extract the common factor from each of the last two rows:
$$
D=\frac12\,(b-a)(c-d)(b-c)(b-d)(a-c)\begin{vmatrix}
2& c+d&2cd\\
0& 1& 2a\\
0&1&2d
\end{vmatrix}.
$$
Now the remaining determinant trivially seen to be $4(d-a)$, so
$$
D=2(b-a)(c-d)(b-c)(b-d)(a-c)(d-a).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is there a general formula for $I(m,n)$? Consider the integral
$$I(m,n):=\int_0^{\infty} \frac{x^m}{x^n+1}\,\mathrm dx$$
For $m=0$, a general formula is $$I(0,n)=\frac{\frac{\pi}{n}}{\sin\left(\frac{\pi}{n}\right)}$$
Some other values are $$I(1,3)=\frac{2\pi}{3\sqrt{3}}$$ $$I(1,4)=\frac{\pi}{4}$$ $$I(2,4)=\frac{\pi}{2\sqrt{2}}$$
For natural $m,n$ the integral exists if and only if $n\ge m+2$.
Is there a general formula for $I(m,n)$ with integers $m,n$ and $0\le m\le n-2$ ?
| We shall compute it in two steps. First, perform the substitution $y = x^n$ in order to get
$$I(m,n) = \int \limits _0 ^\infty \frac {y ^{\frac m n}} {1 + y} \frac 1 n y ^{\frac 1 n - 1} \ \Bbb d y = \frac 1 n \int \limits _0 ^\infty \frac {y ^{\frac {m+1} n - 1}} {1 + y} \ \Bbb d y .$$
Now perform the change $t = \frac y {1+y}$, to obtain
$$I(m,n) = \frac 1 n \int \limits _0 ^1 \frac {\left( \frac t {1-t} \right) ^{\frac {m+1} n - 1}} {1 + \frac t {1-t}} \frac 1 {(1-t)^2} \ \Bbb d t = \frac 1 n \int \limits _0 ^1 t^{\frac {m+1} n - 1} (1-t)^{- \frac {m+1} n} \ \Bbb d t = \frac 1 n B \left( \frac {m+1} n, 1 - \frac {m+1} n \right) = \frac 1 n \frac \pi {\sin \pi {\frac {m + 1} n}} .$$
In the above, $B$ is Euler's Beta function and I have used the known formula $B(x, 1-x) = \frac \pi {\sin \pi x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$ But now, I'm stuck. Solutions are greatly appreciated.
| HINT: You know what $\sin^2\alpha+\cos^2\alpha$ is, so the problem boils down to sorting out $\sin\alpha\cos\alpha$. But
$$(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\;,$$
and you can solve this for $\sin\alpha\cos\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
A problem related to arithmetico-geometric sequence Question:
Find the sum of the series: $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ... = 1+\sum_{n=1}^\infty \frac{(4n-2)}{3^n}$
My doubt:
I have taken $\frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ...$ to be in A.G.P. and calculated the sum using the formula $$S_\infty=\frac{a}{1-r}+\frac{d \cdot r}{(1-r)^2},$$ where $a=\frac23, r=\frac13$ and $d=4$.
I get $S_\infty=4$ and then I add $1$ to it to get the answer $5$.
But my book shows a different method with a different answer ($3$).
Please tell what is wrong with this method.
| It's not a standard AGP. An AGP has the recurrence relation $u_{n+1}=r*u_n+b$, but here that isn't the case. So that's why you can't plug it into the formula for an AGP!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
functional equation of type $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$
If $f:\mathbb{R}-\{-1\}\rightarrow \mathbb{R}$ and $f$ is a differentiable function that satisfies $$f(x+f(y)+xf(y)) = y+f(x)+yf(x)\forall x,y \in \mathbb{R}-\{-1\}\;,$$ Then value of $\displaystyle 2016(1+f(2015)) = $
$\bf{My\; Try::}$ Using partial Differentiation, Differentiate w r to $x$ and $y$ as a constant
$$f'(x+f(y)+xf(y)) \cdot (1+f(y)) = f'(x)+yf'(x)$$
Similarly Differentiate w r to $y$ and $x$ as a constant
$$f'(x+f(y)+xf(y))\cdot (f'(y)+xf'(y)) = 1+f(x)$$
Now Divide these two equation, We get $$\frac{1+f(y)}{(1+x)f'(y)} = \frac{(1+y)f'(x)}{1+f(x)}$$
Now How can i solve it after that, Help required, Thanks
| (Work in progress; I was unexpectedly called away. The aim of this post is to do it without assuming differentiability.)
Roots
To find the roots of $f$: suppose $f(y) = 0$.
Then $f(x) = y+f(x)+y f(x)$ and so $(1+f(x))y = 0$; therefore $f(x) = -1$ for all $x$, or $y=0$.
So we have two possibilities: $f(x) = -1$ for all $x \not = -1$, or $f(0) = 0$.
But if $f(x) = -1$ for all $x \not = -1$, then letting $x=0$ we obtain $$f(-1) = -1$$
which is a contradiction to the undefinedness of $f(-1)$.
(There's a slightly simpler contradiction if we simply suppose there is any $y$ with $f(y) = -1$, as egreg points out.)
Therefore $f$ has exactly one root, and it is at $0$.
(Strictly, we should go back through this, and verify that we never tried to give $f$ the argument $-1$. The easiest way to do this is to let $x$ be some real such that $f(x) \not = -1$, and then just go through the same proof again.)
Using the root
Letting $x=0$, we see that $f(f(y)) = y$ for all $y$.
Letting $y=f(x)$, we obtain $$f(2x+x^2) = 2f(x)+f(x)^2$$
so, if $x=-2$, we get $f(0) = 2 f(-2) + f(-2)^2$; that is, $f(-2) = 0$ or $f(-2) = -2$.
We already know $f$ has exactly one root, so $f(-2) = -2$.
Letting $x = f(y)$, we obtain $$f(2f(y)+f(y)^2) = 2y+y^2$$
so if $f(y) = -2$, we obtain $y = 0$ (contradiction) or $y=-2$.
Therefore $f$ hits $-2$ at exactly one input: namely $-2$.
Letting $y=-2$, then, we obtain $f(-x-2) = -2+f(x)-2f(x)$, so $$f(-x-2) = -f(x)-2$$
and so the behaviour of the function is determined precisely by its behaviour on $x>-1$.
Actually, under the assumption that there is $x$ such that $f(x) = 1$, we obtain $f(x + f(y) + x f(y)) = 1 + 2 y$ and hence (by applying $f$ to both sides) $$x + f(y) + x f(y) = f(1 + 2 y)$$
whence (substituting $z=1+2y$) we get $$f(z) = f\left(\frac{z-1}{2}\right) + x f\left(\frac{z-1}{2}\right) + x$$
so any open interval $(-1, r)$ determines the behaviour of $f$ completely.
TODO: complete this. As LutzL points out, there is more than one function satisfying the recurrence :(
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Deriving the equation of a circle I have two points, $A=(-1,3)$ and $B=(2,7)$. There is a third point $P=(x,y)$.
I have found $|AP|$ and $|BP|$ in terms of $x$ and $y$. These are:
$|AP| = \sqrt{(-1-x)^2 +(3-y)^2}$
$|BP| = \sqrt{(2-x)^2 +(7-y)^2}$
I have shown that the set $S = \{P:|AP| = |BP|\}$ is a straight line and determined the slope and y-intercept by expanding out $|AP| = |BP|$ and simplifying to get:
$y= \frac{-3}{4}x+\frac{43}{8}$.
What I can't derive is an equation for the circle $C= \{P: |AP| = \alpha|BP|\}$. I know it should be of the form $(x-a)^2 + (y-b)^2 = \alpha^2$ but when I try to tackle this algebraically I get stuck.
| $$|AP|^2=(x+1)^2+(y-3)^2$$
$$|BP|^2=(x-2)^2+(y-7)^2$$
The equation of $C$ may be worked out as follows:
$$|AP|=\alpha|BP|$$
$$|AP|^2=\alpha^2|BP|^2$$
$$(x+1)^2+(y-3)^2=\alpha^2((x-2)^2+(y-7)^2)$$
$$x^2+2x+1+y^2-6y+9=\alpha^2(x^2-4x+4+y^2-14y+49)$$
$$(\alpha^2-1)x^2+(\alpha^2-1)y^2-(\alpha^2+\tfrac12)(4x)-(\alpha^2-\tfrac37)(14y)=10-53\alpha^2$$
$$x^2+y^2-\frac{2(\alpha^2+\frac12)}{\alpha^2-1}(2x)-\frac{7(\alpha^2-\frac37)}{\alpha^2-1}(2y)=\frac{10-53\alpha^2}{\alpha^2-1}$$
Let $M=\frac{2(\alpha^2+\frac12)}{\alpha^2-1}$, $N=\frac{7(\alpha^2-\frac37)}{\alpha^2-1}$ and $K=\frac{10-53\alpha^2}{\alpha^2-1}$, then
$$x^2+y^2-2Mx-2Ny=K$$
$$(x-M)^2+(y-N)^2=K+M^2+N^2$$
This last equation of $C$ is now in the canonical form for a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
$$41-4\cdot17=1$$
So the inverse of $17$ is $-4$.
That is, the inverse of $17$ is $37$
Am I right?
| $$17x\equiv 1\pmod{41}\\-24x\equiv -40\pmod{41}\\3x\equiv 5\pmod{41}\\3x\equiv-36\pmod{41}\\x\equiv -12\equiv29\pmod{41}$$
Multiplying $x\equiv29\pmod{41}$ by $14$ also gives a correct solution though I'm not quite sure if it's a coincidence or not.$$\\17x\equiv14\pmod{41}\\-24x\equiv14\pmod{41}\\-12x\equiv 7\pmod{41}\\-12x\equiv-34\pmod{41}\\6x\equiv17\pmod{41}\\6x\equiv-24\pmod{41}\\x\equiv -4\equiv37\pmod{41}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
Explain this equality: $-90^\circ + \tan^{-1}\frac{X_C}{R} = - \tan^{-1}\frac{R}{X_C}$
$$-90^\circ + \tan^{-1}\frac{X_C}{R} \;=\; - \tan^{-1}\frac{R}{X_C}$$
Please explain how both are equal.
| $$\text{Let} \ \ \arctan \frac{1}{x} = \theta$$
$$\frac{1}{x} = \tan\theta$$
$$x = \cot \theta = \tan \left(\frac{\pi}{2} - \theta\right)$$
$$\arctan x = \frac{\pi}{2} - \theta$$
So, $$ \arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
points on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is minimum
Find point on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is, minimum
$\bf{My\; Try::}$ Let $(x,y)$ be any point on the curve $x^2+2y^2=6\;,$ Then we have to
minimize $\displaystyle \left|\frac{x+y-7}{\sqrt{2}}\right|$
Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\left[x^2+\left(\sqrt{2}y\right)^2\right]\cdot \left[1^2+\left(\frac{1}{\sqrt{2}}\right)^2\right]\geq (x+y)^2$$
So $$6\times \frac{3}{2}\geq (x+y)^2\Rightarrow (x+y)^2\leq 3^2\Rightarrow-3 \leq (x+y)\leq 3$$
So We get $$\frac{-3-7}{\sqrt{2}}\leq \frac{x+y-7}{\sqrt{2}}\leq \frac{3-7}{\sqrt{2}}\Rightarrow -5\sqrt{2}\leq \frac{x+y-7}{\sqrt{2}}\leq -2\sqrt{2}$$
and equality hold when $\displaystyle \frac{x}{1} = \frac{2y}{1}$
Now i did not understand how can i calculate $\displaystyle \min \left|\frac{x+y-7}{\sqrt{2}}\right|,$ Help me thanks
| Let $t = x+y$, you've shown $-3 \le t \le 3$, thus the function $f(t) =\dfrac{ (t-7)^2}{2}$ has a graph a parabola with the axis of symmetry at $x = 7$, and it is decreasing on $[-3,3]$, thus the min is $f(3) = 8$. Thus the min distance is $2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to symbolically solve a system of linear equations A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \frac{1}{4}a+\frac{3}{4}b\\
(\frac{3}{4}a+\frac{1}{4}b)^2 & (\frac{1}{2}a+\frac{1}{2}b)^2 &
(\frac{1}{4}a+\frac{3}{4}b)^2\\ \end{array} \right] \cdot
\left[ \begin{array}{c} w_1\\ w_2\\ w_3 \end{array} \right] =
\left[ \begin{array}{c} b-a\\ \frac{b^2-a^2}{2}\\
\frac{b^3-a^3}{3} \end{array} \right] $$ where $a, b$ are
constants? (I'm trying to derive the formula for a 3-point open
Newton-Cotes quadrature rule.) Thanks!
Unfortunately the question was deleted (link), but I think it might be interesting in general.
| One can also make use of the Mathematica programming language and proceed as follows:
where the lines with MatrixForm[$\cdots$] were just to check if the matrices were fine.
The command LinearSolve[matrix, result] finds the solution and the added // MatrixForm is just for pretty printing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that$\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$=$\frac{1-{x^2}}{2x}$ Well, I was trying to find $\tan40$ in terms of $\tan25$=x.
So, I expanded 40 as 25+15 and got the value of $\tan40$ in terms of $x$ as $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$.
Now, I solved $\tan40$ alternatively as
$$\frac{\tan155-\tan115}{1+\tan155\tan115}= \frac{\tan(180-25)-tan(90+25)}{1+\tan(180-25)tan(90+25)}= \frac{-\tan25+\cot25}{1+\tan25{\cot25}}=\frac{1}{2}(-x+\frac{1}{x})=\frac{1-{x^2}}{2x}$$
But, we should have same solution from both the approaches.
Therefore, $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$ must be equals to $\frac{1-{x^2}}{2x}$. But I'm unable to prove it by algebra. So, can you kindly help me?
| When $\frac{1-x^2}{2x}$ is equated to $\frac{x+2−\sqrt{3}}{1+(\sqrt{3}-2)x}$, it gives rise to: $$x^3-3(\sqrt{3}+2)x^2-3x+(\sqrt{3}+2)=0$$
Wolfram gives three roots:$$x≈-0.700208,\,\,\,x≈0.466308,\,\,\,x≈11.4301$$
Also, note that, $\tan{25}≈0.4663076...$
It is not a satisfying "exact" proof, but something to get a sense of things.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction, that $ \sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!}$ If I'm not wrong,
$$\sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!},$$
but I am having trouble proving it by induction...
If $n=1$ the formulas coincide.
Sup $n=k$ is valid: $$\sum_{i=1}^k \frac{i}{(i+1)!}= \frac{1}{(k+1)(k-1)!}.$$
Then if $n=k+1$,
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+1)!} $$
$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+1)!} $$
$$ = \frac{k + (k +1)}{(k+1)!} = \frac{2k + 1}{(k+1)!} $$ I really don't see from this how I should arrive to
$$ \sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \frac{1}{(k+2)(k)!} $$
| Try this in your inductive step:
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+2)!} $$
Notice the denominator is $(k + 2)!$ instead of $(k+1)!$, which makes simplification easier:
$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+2)!} $$
$$ = \frac{(k + 2)k}{(k+2)!} + \frac{k+1}{(k+2)!} = \frac{k^2 + 3k + 1}{(k+2)!} $$
$$ = \frac{(k + 2)(k + 1) - 1}{(k+2)!} = \frac{1}{k!} - \frac{1}{(k+2)!}$$
It looks like your inductive hypothesis is incorrect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Laplace transform of a differential equation: $y'+y=e^{-3t}\cos(2t)$ Given $y(0)=0$, solve
$$y'+y=e^{3t}cos(2t) $$
Steps:
$$sY(s)-y(0)+Y(s)=\frac{s+3}{(s+3)^2+2^2}$$
$$Y(s)(s+1)=\frac{s+3}{(s+3)^2+2^2}$$
$$Y(s)=\frac{s+1}{(s+1)((s+3)^2+2^2)}+\frac{2}{(s+1)((s+3)^2+2^2)}$$
$$Y(s)=\frac{1}{(s+3)^2+2^2}+\frac{\textbf{2}}{\textbf{(s+1)((s+3)^2+2^2)}}$$
Now I can reverse transform the first fraction, may I know how to deal with bolded fraction?
| Hint. You may use a partial fraction decomposition to get
$$
\frac{2}{(s+1)((s+3)^2+2^2)}=\frac{1}{4(s+1)}-\frac14\cdot\frac{(s+3)}{(s+3)^2+2^2}-\frac12\cdot\frac{1}{(s+3)^2+2^2}
$$ then each term can be classically reversed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Dice problem for the n space There is a 6-faced dice and every time you roll it, you move forward the same number of spaces that you rolled. If the finishing point is “n” spaces away from the starting point, how many possible ways are there to arrive exactly at the finishing point?
| Your questions deals with the number of ["compositions"](https://en.wikipedia.org/wiki/Composition_(combinatorics) of an integer.
Although there is almost surely no closed form formula for it, here is a general technique for obtaining results for a given $n$ by using generating functions.
I will present it through an example: assume you want to reach position $n=12$ in $k=3$ steps, doing for example : $6+5+1$ or $5+6+1$ or $6+4+2$, etc. A hand examination show that there are 28 of them.
Let us now consider the following polynomial expansion:
$$p(x)^3 \ \ \ \ \ \ \text{with} \ \ \ \ \ p(x):=x+x^2+x^3+x^4+x^5+x^6$$
Using a Computer Algebra System like Mathematica, we obtain:
$$1 + 3\,x + 6\,x^2 + 10\,x^3 + 15\,x^4 + 21\,x^5 + 28\,x^6 + 33\,x^7 +
36\,x^8 + 37\,x^9 + 36\,x^{10} + 33\,x^{11} + 28\,x^{12} +
21\,x^{13} + 15\,x^{14} + 10\,x^{15} + 6\,x^{16} + 3\,x^{17} +
x^{18}.$$
The number of compositions of $n=12$ into $k$ parts is found directly as the coefficient 28 of $x^{12}$.
Explanation: The general term in the development of
$$(x+x^2+x^3+x^4+x^5+x^6)(x+x^2+x^3+x^4+x^5+x^6)(x+x^2+x^3+x^4+x^5+x^6)$$
is $x^{i+j+k},$ obtained by taking a $x^i$, a $x^j$, a $x^k$ in each of the three parentheses ; among all these $x^{i+j+k}$, keeping only those whose exponent $i+j+k$ is exactly $12$ amounts to "sweep" all "compositions" of $n=12$ into $k=3$ parts:
$$x^{6+5+1}+x^{5+6+1}+x^{6+4+2}+...$$
Keeping $n=12$ fixed, you now do these operations for all possible $k$, i.e., for all $k=3,4,...12$ (if course $k=1$ and $k=2$ are impossible).
Thus the answer is: the coefficient of $x^{12}$ in:
$$q(x)=p(x)^3+p(x)^4+\cdots+p(x)^{12}$$
which can also be written:
$$p(x)^3(1+p(x)+p(x)^2+\cdots+p(x)^9)=p(x)^3\dfrac{1-p(x)^{10}}{1-p(x)}$$
What has worked for $n=12$ will work of course for any other $n$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to directly expand the determinant before greatly simplifying it).
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}= (ab+bc+ca)^3$$
I tried everything:
$R_1\to R_1+R_2+R_3$ and similar transformations to extract that $ab+bc+ca$ term, but to no avail. $C_2\to C_1+C_2$ and $C_3\to C_3+C_1$ seemed to be a good lead, but I couldn't follow up.
How can I solve this question?
| $A:=\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\enspace\enspace\enspace$ Shift: $\enspace C_2\to C_1$, $C_3\to C_2$, $C_1\to C_3\enspace$ => $\enspace B$
$B:=\begin{vmatrix}
b^2+bc & c^2+bc & -bc\\
-ac & c^2+ac & a^2+ac \\
b^2+ab & -ab & a^2+ab
\end{vmatrix}$
$C:=\begin{vmatrix}
0 & 0 & (ab+ac+bc)^2\\
(ab+ac+bc)^2 & 0 & 0 \\
0& (ab+ac+bc)^2 & 0
\end{vmatrix}$
It's $|A|\cdot|B|=|A\cdot B|=|C|$ with $|A|=|B|$ (Sarrus rule without calculating, only a compare).
=> $|A|=\sqrt{|C|}=(ab+ac+bc)^3$
Addition:
In order to calculate the sign of the square root, it's sufficient to put in a value, e.g. $(a;b;c):=(1;1;0)$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y \sqrt{x z},\\
x z +y z \geq 2 z \sqrt{xz}.
\end{gather*}
Add and get
$$
xy+xz+yz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}.
$$
By condition
$$
xyz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}
$$
Dividing by $\sqrt{xyz}$ we obtain the inequality.
Question. Are there another ways to prove it?
| My another solution inspired by Gordon's solution.
Suppose that $x \leq y \leq z$ Then $\sqrt{x} \leq \sqrt{y} \leq \sqrt{z}$ and
$$
\frac{1}{\sqrt{x}} \geq \frac{1}{\sqrt{y}} \geq \frac{1}{\sqrt{z}}.
$$
Now by rearrangement inequality we get
$$
1 \geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{xz}} +\frac{1}{\sqrt{yz}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How does $ab+ac+bc =0$ imply that $a+b+c = \pm 1$? Show that $$ A=
\begin{bmatrix}
a & c & b \\
b & a & c \\
c & b & a \\
\end{bmatrix}
$$ is orthogonal if and only if $a^2+b^2+c^2=1$ and $a+b+c= \pm 1$.
For a matrix to be orthogonal it must satisfy $A^T A^{-1} = I$ where $I$ is the identity matrix.
I calculated $$A^T A^{-1} =\begin{bmatrix}
a^2+b^2+c^2 & ab+ac+bc & ab+ac+bc \\
ab+ac+bc & a^2+b^2+c^2 & ab+ac+bc \\
ab+ac+bc & ab+ac+bc & a^2+b^2+c^2 \\
\end{bmatrix} $$
So for $A^T A^{-1} = I$ we must have that $$ a^2+b^2+c^2 = 1$$ and $$ab+ac+bc =0 $$
So then I showed that $ a^2+b^2+c^2 = 1$, but through a lot of rough work I can't seem to make $ab+ac+bc =0 $ imply that $a+b+c= \pm 1$. Does anyone have any hints?I feel like i've just about tried every way to sub something into something else!
| The comments should be enough,
but
if $a^2+b^2+c^2 = 1$
and
$ab+ac+bc = 0$,
then
$(a+b+c)^2
=a^2+b^2+c^2+2(ab+ac+bc)
=1
$
so
$a+b+c = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what kind of test should I use to determine if this sequence converge or diverge? what kind of test should I use to determine if this sequence converge or diverge?
$$\sum_{n=1}^\infty \left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right)$$
| By comparison test $$sin\frac{1}{n}-sin \frac{1}{n+1}=\\2sin(\frac{1}{n}-\frac{1}{n+1})cos(\frac{1}{n}+\frac{1}{n+1})=\\2sin(\frac{1}{n(n+1)})cos(\frac{2n+1}{n(n+1)})\\\leq2sin(\frac{1}{n(n+1)}) \\2\frac{1}{n(n+1)} \leq2 \frac{1}{n^2}\\$$then
$$\sum sin\frac{1}{n}-sin \frac{1}{n+1} \leq \sum \frac{2}{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
\hline
x^2 & \\
\hline
& 3 \\
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
\hline
x^2 & 1x \\
\hline
3x& 3 \\
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
$$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
| X(X+1)(X+3)
Therefore (X+1)(X+3) = X^2+4X+3.
Therefore multiplying this by X you get X^3+4X^2+3X.
So yes well done that is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
finding the limit of an expression with roots I need to find the limit of the expression:
$$
\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}
$$
It is given that $b>2$. In the class we were encouraged to use the method of multiplying both the numerator and the denominator with the numerator when the limit can't be found simply by using limit arithmetic. Hence:
$$
\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2} =\\
=\lim_{x\to b}\frac{(\sqrt{b-2} - \sqrt{x-2})(\sqrt{b-2} - \sqrt{x-2})}{(x^2-b^2)(\sqrt{b-2} - \sqrt{x-2})} = \\
= \lim_{x\to b}\frac{b+x-4 - 2\sqrt{b+x-4}}{(x-b)(x+b)(\sqrt{b-2} - \sqrt{x-2})}
$$
I don't see though how from this point this can be simplified. We could factor out the $\sqrt{b+x-4}$ in the numerator but this doesn't really help. Wolfram says the answer is $\frac{1}{4b\sqrt{b-2}}$
| Multiply by $\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$
$$\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$$
$$\lim_{x\to b}\frac{b-x}{(x-b)(x+b)(\sqrt{b-2} + \sqrt{x-2})}=\frac{-1}{(b+b)(\sqrt{b-2}+\sqrt{b-2})}=\frac{-1}{4b\sqrt{b-2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side?
| You may note that $3+2\sqrt{2}=(1+\sqrt{2})^2$. Hence, we have
$$
\sqrt{2}\sqrt{3+2\sqrt{2}}-1=\sqrt{2}(1+\sqrt{2})-1=\sqrt{2}+2-1=\sqrt{2}+1,
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equilateral.
$\bf{My\; Try::}$ We can write it as $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1$$
So $$\left(\frac{a+b-c}{a}\right)^{a}\cdot \left(\frac{b+c-a}{b}\right)^{b}\cdot \left(\frac{c+a-b}{c}\right)^{c}\geq 1$$
So $$(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$$
Now how can i solve after that, Help required, Thanks
| Another solution:
Prove: $$\left ( a+ b- c \right )^{a}\left ( b+ c- a \right )^{b}\left ( c+ a- b \right )^{c}\leq a^{a}b^{b}c^{c}$$
By weight power mean:
$$\sqrt[a+ b+ c]{\left ( \frac{a+ b- c}{a} \right )^{a}\left ( \frac{b+ c- a}{b} \right )^{b}\left ( \frac{c+ a- b}{c} \right )^{c}}\leq \frac{1}{a+ b+ c}\left ( a. \frac{a+ b- c}{a}+ b. \frac{b+ c- a}{b}+ c. \frac{c+ a- b}{c} \right )= 1$$
i.e.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
UK 1998, Show that $hxyz$ is a perfect square. Problem: [UK 1998] Let $x,y$ and $z$ be positive integers such that $$\frac{1}{x}-\frac{1}{y}=\frac{1}{z}.$$ Let $h=(x,y,z).$ Prove that $hxyz$ is a perfect square.
My Attempt: Of course $y>x$ and therefore let $y=x+a$ for some $a>0.$ Then $$z=\frac{x(x+a)}{a}.$$ Also $$h=(x,x+a,z)=(x,a,z).$$ Now $a|x(x+a)$ since $z$ is a positive integer. Let $p$ be a prime divisor of $a$ then $p|x$ or $p|x+a$. But if $p|x\Rightarrow p|x+a$ and similarily if $p|x+a\Rightarrow p|x.$ In any case we deduce that $a|x$ and $a|x+a$ also. Thus $$z=\frac{x(x+a)}{a}=\frac{a^2k_1}{a}=ak_1$$ and $x=ak_2.$ Thus $h=a.$ The expression $hxyz=a(az)(z)=(az)^2$ which is indeed a whole square.
I would like to know whether this proof is correct or not since the argument provided in the solution is quite different from the one I've written in this question.
| Let's try your argument with $x = 15; y = 40; z = 24$ where $\frac 1{15} - \frac 1{40}= \frac{8}{8*15} - \frac{3}{3*40} = \frac 5{5*24} = \frac 1 {24}$ and $h = (15,40,24) = 1$ and $hxyz = 1*3*5*8*5 *2*8 = (3*5*8)^2$.
You argue $40 > 15$ and $a = 40 -15 = 25$.
So $z = 24 = \frac {15*(15 + 25)}{25}$ which is indeed true.
You argue that $h = (15,40,24) = (15,25,24)$ which is true.
You argue that if $p = 5$ a prime factor of $a = 25$ than $5|15*(15+ 25)$ and that as $5|25$ then $5|x(x + 25) \iff 5|x = 15$ which is true.
Then you conclude that since $5|15$ then $25|15$ which is not true. We can only conclude the single power products of the prime factors divide $x$; not any powers of any prime.
You argue $z = \frac{x(x+a)}{a} = \frac {a^2k_1}{a} = ak^2$ or $24 = \frac{15(15 + 25)}{25} = \frac {25^2*\frac 35(\frac 35 +1)}{25} = 25*\frac 35\frac 85 = 25*\frac {24}{25}$. In which case $k_1=\frac {24}{25}$ is not an integer.
Likewise $15 = 25*\frac 35$ but $\frac 35$ is not an integer. $25$ is not a common divisor to $15, 24, $ and $h \ne 25$.
Actually $h = a > 1$ should have set off huge red flags. $h = (x,y,z)$ so this means $x,y,z$ are never relatively prime? Well, what happens if we factor $h$ out of them all? We have $\frac 1{x'} + \frac 1{y'} = \frac 1{z'}$ where $y' = x + 1$ and $z' = x'(x'+1)$? Is that really the only possibility?
I think it can be fixed but it doesn't work as is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \frac{x^2\sqrt{1+\frac{\tanh(x)}{x}+\frac{1}{x^{2}}}-x^2(1+\frac{1}{x})}{x(1+\frac{1}{x})}$$
and that won't cancel with the x which I have on the bottom so the limit just blows up. Did I make a mistake somewhere?
The limit is apparently $\frac{-1}{2}$ but i'm not sure how that's the case. Thanks.
| $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
$$=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x+1)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x^2+2x+1)}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^4-2x^3-x^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{(x+1)\left(x^2\sqrt{1+\frac {\tanh(x)}x+\frac 1 {x^2}}+x^2+x\right)}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{\left(x^3\sqrt{1+\frac {\tanh(x)}x+\frac 1 {x^2}}+x^3+x^2\right)+\left(x^2\sqrt{1+\frac {\tanh(x)}x+\frac 1 {x^2}}+x^2+x\right)}$$
$$=\lim_{x\to\infty} \frac{\tanh(x)-2}{\left(\sqrt{1+\frac {\tanh(x)}x+\frac 1 {x^2}}+1+\frac 1 x\right)+\left(\frac 1 x\sqrt{1+\frac {\tanh(x)}x+\frac 1 {x^2}}+\frac 1 x+ \frac 1 {x^2}\right)} = -\frac 1 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is there a closed form of the sum $\sum _{n=1}^x\lfloor n \sqrt{2}\rfloor$ I need to find the n-th partial sum of:
$$\sum _{n=1}^x\lfloor n \sqrt{2}\rfloor$$
Or this sum of a Beatty sequence.
I tried to expand as the following:
$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x\sum _{k=0}^{\infty}\frac{\sin \left(2 \pi \sqrt{2} k\ n\right)}{k}$$
$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x \arctan (\tan( \frac{\pi - 2 \pi \sqrt{2} n}{2}))$$
For $ n ∈ \{0,\frac{1}{\sqrt{2}}\}$
$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x \frac{\pi - 2 \pi \sqrt{2} n}{2}$$
But I still have the last term that oscillates around some value and I cant figure out how to find the result without doing the sum up to x. if there is no closed form, is it possible to find it for special values of x? Any hint is appreciated.
| Since $\lfloor \sqrt 2 k \rfloor$ and $\lfloor (2+\sqrt 2) k \rfloor$ are complementary Beatty sequences, they partition the integers and we can find a recurrence relation for their sum. To sum the sequence $\lfloor \sqrt 2 k \rfloor$ we can use a well-known formula to add the consecutive integers
$$
\begin{align}
\sum_1^nk=&1+2+3+..+k+..+n\\
=&\frac{n(n+1)}2
\end{align}
$$
and subtract the terms that should be missed. That is,
$$
\begin{align}
&(1+2+4+..+\lfloor \sqrt 2 k \rfloor+..+\lfloor \sqrt 2 n \rfloor)\\
=&(1+2+3+..+k+..+\lfloor \sqrt 2 n \rfloor)\\
&-(3+6+..+\lfloor (2+\sqrt 2) k \rfloor+...+\lfloor (2+\sqrt 2) {n'} \rfloor)\\
=&(1+2+3+..+k+..+n^*)\\
&-2(1+2+..+k+...+n')\\
&-(1+2+4+..+\lfloor \sqrt 2 k \rfloor+...+\lfloor \sqrt 2 {n'} \rfloor)\\
\end{align}
$$
where $n'=\lfloor (\sqrt 2-1) n \rfloor$ and $n^*=\lfloor \sqrt 2 n \rfloor$, giving us
$$\sum_{k=1}^n\lfloor \sqrt 2 k \rfloor=\frac {n^*(n^*+1)}2 -n'(n'+1)-\sum_{k=1}^{n'} \lfloor \sqrt 2 k \rfloor$$
(Note that $n^*=n'+n$). Since the recurrence relation reduces the number of terms by a factor of $1+\sqrt 2$ on each iteration, we might be able to analyse this relation when $n$ is (near a multiple of) a power of $1+\sqrt2$ and come up with a closed formula for the sum. Let us define some integers for this purpose. Let
$$
\begin{align}
p_m&=\frac{(1+\sqrt 2)^m-(1-\sqrt 2)}{2\sqrt 2}^m\\
\text{and }q_m&=\frac{(1+\sqrt 2)^m+(1-\sqrt 2)}{2}^m
\end{align}
$$
Since
$$
\begin{align}
&p_0=0,\\
&p_1=q_0=q_1=1,\\
&p_m=2p_{m-1}+p_{m-2}\\
\text{and } &q_m=2q_{m-1}+q_{m-2},\\
\end{align}
$$
then $p_m$ and $q_m$ are integers for each integer $m$.
As an example, we will take $n=q_m$ so
$$
\begin{align}
n'&=\lfloor (\sqrt2-1)n\rfloor\\
&=\lfloor \frac 12( \sqrt2-1)(1+\sqrt2)^m+\frac 12(\sqrt2-1)(1-\sqrt2)^m\rfloor\\
&=\lfloor \frac 12(1+\sqrt2)^{m-1}-\frac 12(1+\sqrt2)(1-\sqrt2)^m+\sqrt2(1-\sqrt2)^m\rfloor\\
&=\lfloor \frac 12(1+\sqrt2)^{m-1}+\frac 12(1-\sqrt2)^{m-1}+\sqrt 2(1-\sqrt2)^{m}\rfloor\\
&=q_{m-1}+\lfloor \sqrt 2(1-\sqrt2)^{m}\rfloor\\
&=q_{m-1}+\begin{cases}
1, & \text{if $m=0$} \\
-1, & \text{if $m$ is odd} \\
0, & \text{if $m>1$ and is even}
\end{cases}
\end{align}
$$
The recurrence relation is now amenable to analytical tools because we have formulas that do not involve the floor function. Take $m>1$ for simplicity. Checking both even and odd $m$, we obtain the same recurrence formula in either case so
$$
\begin{align}
\sum_1^{q_m}\lfloor \sqrt2k\rfloor=&\frac{q_m(q_m+1)-q_{m-1}(q_{m-1}+1)}2\\
&+q_{m}q_{m-1}-\sum_1^{q_{m-1}}\lfloor \sqrt2n\rfloor
\end{align}
$$
This allows us to show by induction on $m$ that
$$2\sum_{k=1}^{q_m}\lfloor \sqrt 2 k \rfloor=p_{2m}+q_{m-1}+(-1)^{m-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
For $y = 2f\left(0.5x+4\right)-3$, derive the equation of: $ f(x)=2x^2-3x+4. $
For $y = 2f\left(0.5x+4\right)-3$,
derive the equation of:
$$
f(x)=2x^2-3x+4.
$$
I don't really understand what this question is asking for. Any input would be greatly appreciated.
| I think that the (admittedly poorly worded) problem is asking for you to express $y$ in terms of $x$ given that $\displaystyle y=2\ f\left(\frac x 2 + 4\right)-3$ and $f(x)=2x^2-3x+4$.
In that case:
\begin{align}
y&=2\ f\left(\frac x 2 + 4\right) - 3\\
&=2\left(2\left(\frac x 2 + 4\right)^2 - 3 \left(\frac x 2 + 4\right)+4\right)-3\\
&=\dots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving algebraic operations for this simple problem I cannot wrap my heard around solving this:
$$\mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}=2\mu\left[1-\left(\frac{1}{2}\right)^{n+1}\right]$$
I have done this instead:
\begin{align}
\mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}&=\mu\frac{1}{\frac{1}{2}}-\frac{(\frac{1}{2})^{n}(\frac{1}{2})^{1}}{\frac{1}{2}}\\
&=2\mu-\frac{1}{2}^{n}
\end{align}
What have I done wrong?
| thanks guys!
What about this one
\begin{align}
\mu\frac{1-(\frac{1}{4})^{n+1}}{\frac{3}{4}}&=\mu\frac{1}{\frac{3}{4}}-\frac{(\frac{1}{4})^{n}(\frac{1}{4})^{1}}{\frac{3}{4}}\\
\end{align}
I don't know how to proceed from this other than
\begin{align} \frac{4}{3}\mu-\mu(\frac{3}{4})^{n+1} \end{align}
which is obviosly wrong. As the result should be
\begin{align} \frac{4}{3}\mu[1-(\frac{1}{4})^{n+1}] \end{align}
Thanks
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On the integral $\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}$ and the plastic constant We have,
$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$$
$$P=\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\cdots}}}$$
with golden ratio $\phi$ and plastic constant $P$. If,
$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=\int_0^1 \frac{2}{1+\sqrt{1+4x}}dx=\frac{2}{\phi}-\ln \phi$$
given in this answer, would it follow that the integral,
$$\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}=\,?$$
is expressible in terms of the plastic constant?
| HINT:
Let $\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}=y\implies x+y=y^3\iff x=y^3-y$
$dx=3y^2-1$
$x=1\implies y\approx 1.3247179572447458$
$$\int \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}=\int\dfrac{3y^2-1}y\ dy=\dfrac{3y^2}2-\ln|y|+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Reduction formula for integral of $\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx $ The following reduction formula is taken from http://www.sosmath.com/tables/integral/integ15/integ15.html:
$$\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx = -\frac{\sqrt{ax^2+bx+c}}{cx} - \frac{b}{2c} \int \frac{1}{x \sqrt{ax^2+bx+c}} dx$$
*
*I've been trying to derive this reduction formula myself, but without success. Can anyone point me in the right direction?
*Are there similar formulae for higher exponents of the $x$ in the denominator? Or even better, for general integer exponents $n$?
| One may observe that
$$
\begin{align}
\left(-\frac{\sqrt{ax^2+bx+c}}{cx}\right)'&=\frac{\sqrt{ax^2+bx+c}}{c x^2}-\frac{2ax+b}{2cx\sqrt{ax^2+bx+c}}
\\\\&=\frac{2(ax^2+bx+c)-(2ax+b)x}{2cx^2 \sqrt{ax^2+bx+c}}
\\\\&=\frac{bx+2c}{2cx^2 \sqrt{ax^2+bx+c}}
\\\\&=\frac{b}{2c}\cdot\frac{1}{x \sqrt{ax^2+bx+c}}+\frac{1}{x^2 \sqrt{ax^2+bx+c}}
\end{align}
$$ which yields the first result.
This might be generalized to get
$$
\begin{align}
\left(-\frac{\sqrt{ax^2+bx+c}}{cx^{\color{red}{n}+1}}\right)'&=\frac{a\cdot \color{red}{n}}{cx^{\color{red}{n}}\sqrt{ax^2+bx+c}}+\frac{b}{2c}\cdot\frac{(2\color{red}{n}+1)}{x^{\color{red}{n}+1} \sqrt{ax^2+bx+c}}+\frac{\color{red}{n}+1}{x^{\color{red}{n}+2}\sqrt{ax^2+bx+c}}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
| What you have looks like
$$x^8+x^6-4x^4+x^2+1$$
Which on rearranging becomes $$(x-1)^2(x+1)^2(x^4+3x^2+1)$$
Since the terms in the product are greater than or equal to zero so the product itself is greater than or equal to zero. i.e
$$x^8+x^6-4x^4+x^2+1\ge0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Finding the roots, with multiplicity, of a polynomial with compound angle formula My question is to find the roots, counted with multiplicity, of the polynomial equation
$16x^5-20x^3+5x-1=0$ using the compound angle formula $\sin\left(5\theta\right)=16\sin^5\theta-20\sin^3\theta+5\sin\theta$
So after substituting $x=\sin\theta$, I get to the equation
$\sin(5\theta)=1$
Then I get an infinitude of $\theta$ values, which when I find the sines of these values, all correspond to the distinct solutions $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(-\frac{3\pi}{10}\right)$.
What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.
Also if possible, is there a way to solve this polynomial using the given compound angle formula without the need to find the distinct roots and then determine the ones which repeat?
This is because the solutions to this question say $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$ without any reasoning, which makes me suspect I am unaware of some related theorem.
| As already explained, the solutions are:
$$x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$$
Because,
$$\sin(5\theta)=1 \Rightarrow \theta=\frac{1}{5}\left(\frac{\pi}{2}+2k\pi\right)$$
and setting $k=0,1,2,3,4$ we get all five solutions.
But $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)$ and $\sin\left(\frac{13\pi}{10}\right)=\sin\left(\frac{17\pi}{10}\right)$.
So the five roots (with multiplicity) are
$$1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right)$$.
Let's explain better why if make any other choice of $k$ we will always find an angle congruent to one of those in $S=\{\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10\}$.
We can write $k=5n+r$ with $r \in \{0,1,2,3,4\}$ and $n \in \Bbb Z$. Replacing that at the expression of $\theta$ we get:
$$\theta=\frac{\pi}{10}+\frac{2k\pi}{5}=\frac{\pi}{10}+\frac{2(5n+r)\pi}{5}=\frac{\pi}{10}+2n\pi+\frac{2r\pi}{5} \equiv \frac{\pi}{10}+\frac{2r\pi}{5} $$
The last equivalence means that the angle $\frac{\pi}{10}+2n\pi$ is congruent to $\frac{\pi}{10}$. That means that both angle stop at the same point on the trigonometric circle and then if we apply any trigonometric function at those angles we will get the same value. The algebric value are different but geometricaly they represent the same point at the circle.
It means that if we want to find the different solution for $\sin\theta$ is enough to take
$$\frac{\pi}{10}+\frac{2r\pi}{5}$$
with $r=0,1,2,3,4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Radius of convergence and convergence sets of $\sum\limits_n\frac{2n+1}{(n-1)^2}x^n$ and $\sum\limits_n(-1)^n(\sqrt{n+1}-\sqrt{n})x^n$ I want to find the radius of convergence of the following series and the set of $x\in \mathbb{R}$ in which the series converge.
*
*$$\sum_{n=2}^{\infty}\frac{2n+1}{(n-1)^2}x^n$$
*$$\sum_{n=0}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})x^n$$
To find the radius of convergence we have to compute the limit $\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}$.
Do we do something elese at the first case where the sum starts from $2$ and not from $0$ ?
I have done the following:
*
*$$\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}=\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{2n+1}{(n-1)^2}x^n\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{2n+1}{(n-1)^2}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{2n+1}{n^2-2n+1}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{\frac{2}{n}+\frac{1}{n^2}}{1-\frac{2}{n^2}+\frac{1}{n^2}}\right|}$$
*$$\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}=\lim_{n\rightarrow \infty}\sqrt[n]{\left|(-1)^n(\sqrt{n+1}-\sqrt{n})x^n\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\sqrt{n+1}-\sqrt{n}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{1}{\sqrt{n+1}+\sqrt{n}}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{1}{\sqrt{n+1}+\sqrt{n}}\right|}$$
$$$$
Is this correct so far? How could we continue?
| We can calculate the radius of convergence by the formula too: $R=lim|\frac{a_{n}}{a_{n+1}}|$, where $a_{n}$ is the $n^{\text{th}}$ term of the given series.
In your first problem, $a_n=\frac{2n+1}{(n-1)^2}$ and $a_{n+1}=\frac{2(n+1)+1}{n^2}$. So $\frac{1}{R}=lim\frac{(2n+3)(n-1)^2}{(2n+1)n^2}=lim\frac{(1+\frac{3}{2n})(1-\frac{1}{n})}{(1+\frac{1}{2n})}=1$. Hence Radius of convergence is $1$.
Similarly, we can calculate the second problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve for all real values of $x$ and $y$ Solve the system of equations for all real values of $x$ and $y$
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$
I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5}}\le y \le {\frac 45}$}
I don't know what to do next.
| Let $z=x + iy \in \mathbb{C}$, now for $|z|^2 = x^2+y^2 \not = 0$ have that:
$$\left(5x + \frac{5x}{x^2+y^2}\right) + i\left(5y - \frac{5y}{x^2+y^2}\right) = 12 + 4i$$
$$5(x+iy) + \frac{5(x-iy)}{x^2+y^2} = 12 + 4i$$
$$5(x+iy) + \frac{5}{x+iy} = 12 + 4i$$
$$5z + \frac{5}{z} = 12 + 4i$$
$$5z^2 - (12+4i)z + 5 = 0$$
Solving this complex equation we get $z_1=\frac 25 - \frac 15i$ and $z_2 =2+i$, corresponding to the solution $(x,y) = \left\{\left(\frac 25, - \frac 15 \right), (2,1)\right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\int \frac{x dx}{x^4 + 6x^2 + 5}$? $$\int \frac{x dx}{x^4 + 6x^2 + 5}$$
How to evaluate this integral ?
| $$\int \frac { xdx }{ x^{ 4 }+6x^{ 2 }+5 } =\frac { 1 }{ 2 } \int { \frac { d{ x }^{ 2 } }{ { \left( { x }^{ 2 }+3 \right) }^{ 2 }-4 } } =\frac { 1 }{ 2 } \left[ \int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+5 \right) } } \right] =\frac { 1 }{ 8 } \left[ \int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) } -\int { \frac { d{ x }^{ 2 } }{ \left( { x }^{ 2 }+5 \right) } } } \right] =\\ =\frac { 1 }{ 8 } \left[ \int { \frac { d\left( { x }^{ 2 }+1 \right) }{ \left( { x }^{ 2 }+1 \right) } -\int { \frac { d\left( { x }^{ 2 }+5 \right) }{ \left( { x }^{ 2 }+5 \right) } } } \right] =\frac { 1 }{ 8 } \ln { \left| \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }+5 } \right| } +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| Let's look at
the area below the part of the curve
from
$(0, 1)$ to $1, 1)$
and the line $y = x$.
The total area is
8 times this.
The area below
$y = x$
is obviously
$1/2$.
If
$x^4+y^4=x^2+y^2
$,
$y^4-y^2
=x^2-x^4
$
or,
trying to be clever,
$y^4-y^2+1/4
=x^2-x^4-1/4+1/2
$
or
$(y^2-1/2)^2
=1/2-(x^2-1/2)^2
$
or
$y^2
=\frac12+\sqrt{\frac12-(x^2-\frac12)^2}
$
so that
$y
=\sqrt{\frac12+\sqrt{\frac12-(x^2-\frac12)^2}}
$.
At the moment,
I don't know how to integrate this.
But,
Wolfy numerically says that
$\int_0^1 y dx
\approx 1.05536
$,
so the area between this
ans $y=x$
is about
$0.55536
$
so that
the total area is about
$8*0.55536
=4.44288
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find an equation for a tangent line to the curve $x^2- y^2=5$ that passes through the point $(1, 1)$. Find an equation for a tangent line to the curve $$ x^2 - y^2 = 5$$ that passes through the point $(1, 1)$.
I realize that I have to use implicit differentiation $$2x - 2y \frac {dy}{dx} = 0$$
$$\frac {dy}{dx} = \frac xy$$
However I do not know how to relate this to the point in order to find the equation of the line that passes through the point, even though the point isn't (presumably) on the curve.
Thank you in advance for the help.
| Here's an approach without (implicit) differentation.
A line passing through $(1,1)$ and with slope $m$ has an equation of the form:
$$y=m(x-1)+1 \iff y = mx-m+1$$
The points of intersection of this line and the hyperbola $x^2-y^2=5$ are the solutions to the following system:
$$\left\{\begin{array}{l}
y = mx-m+1 \\
x^2-y^2=5
\end{array}\right.$$
Substitution of the first into the second equation yields a quadratic equation in $x$:
$$x^2-(mx-m+1)^2=5 \iff \left( 1-m^2 \right)x^2 + \left(2 m^2- 2 m \right) x - m^2 + 2 m - 6 = 0 $$
The line is tangent to the hyperbola if the discriminant is $0$, so:
$$\begin{align}\left(2 m^2- 2 m\right)^2-4\left(1-m^2\right)\left(- m^2 + 2 m - 6 \right) = 0 & \iff -16 m^2 - 8 m + 24 = 0 \\
& \iff 2m^2 + m -3 = 0 \\
& \iff \left( m-1 \right)\left(2m+3 \right) = 0
\end{align}$$
This gives you two slopes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Which would you rather have? Q. Which would you rather have, a piece of an 8-inch pie that's been cut into sixths or a piece of a 10-inch pie that's been cut into eights?
A. This is a problem involving sectors. One-sixth of a pie is $\frac {1}{6}$ of $ 2\pi$ radians. The measure of the central angle is $\frac{1}{6} \cdot 2\pi = \frac{\pi}{3}$. An 8-inch pie has a 4-inch radius. Putting the angle and radius into the formula, for the area of a sector, $A= \frac{1}{2} \cdot \frac{\pi}{3} \cdot 4^2 = \frac{16\pi}{6} = \frac{8\pi}{3} \approx 8.38$ square inches. One-eighth of a pie is $\frac{1}{8}$ of $2\pi$ radians. The measure of the angle is $\frac{1}{8} \cdot 2\pi = \frac{2\pi}{8} = \frac{\pi}{4}$. A 10-inch pie has a 5-inch radius. Putting the angle and radius into the formula, for the area of a sector, $A= \frac{1}{2} \cdot \frac{\pi}{4} \cdot 5^2 = \frac{25\pi}{8} \approx 9.82$ square inches. There isn't too much difference, but it looks like the smaller part of the bigger pie has the larger piece, in terms of area.
My question:
Is how do one, get, $2\pi$ radians?
| Area per piece of 8 inch pie = ${1\over 6} \times \pi (4)^2 = 8.38$
Area per piece of 10 inch pie = ${1\over 8} \times \pi (5)^2 = 9.82$
So the we will take the piece of 10 inch pie. $\ddot \smile$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinants question with factorials $$A= \begin{pmatrix}
n! & (n+1)! & (n+2)!\\
(n+1)! & (n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)!
\end{pmatrix}$$
And $D=\det(A)$,
We have to prove that $D/(n!)^3 - 4$ is divisible by $n$.
I did it by simplifying the determinant using some row and column operations and finally ended up in getting a cubic polynomial in $n$. Then I subtracted 4 from the polynomial and got the result.
Is there any other way I could prove the result quicker and in a better manner?
| $$
\begin{align}
D&=\det\begin{bmatrix}
n!&(n+1)!&(n+2)!\\
(n+1)!&(n+2)!&(n+3)!\\
(n+2)!&(n+3)!&(n+4)!
\end{bmatrix}\\
&=n!(n+1)!(n+2)!
\det\small\begin{bmatrix}
1&1&1\\
n+1&n+2&n+3\\
(n+1)(n+2)&(n+2)(n+3)&(n+3)(n+4)\\
\end{bmatrix}\tag{1}\\
&=n!(n+1)!(n+2)!
\det\begin{bmatrix}
1&1&1\\
0&1&2\\
0&2n+4&4n+10\\
\end{bmatrix}\tag{2}\\
&=n!(n+1)!(n+2)!
\det\begin{bmatrix}
1&1&1\\
0&1&2\\
0&0&2\\
\end{bmatrix}\tag{3}\\[12pt]
&=2n!(n+1)!(n+2)!\tag{4}
\end{align}
$$
Explanation:
$(1)$: divide columns by $n!$, $(n+1)!$, $(n+2)!$ respectively
$(2)$: subtract $n+1$ times row $1$ from row $2$
$\phantom{\text{(2):}}$ subtract $(n+1)(n+2)$ times row $1$ from row $3$
$(3)$: subtract $2n+4$ times row $2$ from row $3$
$(4)$: evaluate the determinant
Therefore,
$$
\begin{align}
\frac{D}{n!^3}-4
&=2(n+1)^2(n+2)-4\\[6pt]
&=n\left(2n^2+8n+10\right)\tag{5}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Evaluate integral $\int_{0}^{2\pi}\frac{1}{1+\cos^2(\theta)}d\theta$ using contours I am trying to evaluate by using contour integration:
$$\int_{0}^{2\pi}\frac{1}{1+\cos^2(\theta)}d\theta$$
From "Fundamentals of Complex Analysis," I have the following two formulas:
$$d\theta = \frac{dz}{iz}, \cos(\theta)=\frac{1}{2}(z+\frac{1}{z})$$ which allow me to rewrite the integral as follows:
$$\int_{C_1}\frac{dz}{iz(1+\frac{1}{4}(z+\frac{1}{z})^2)}=\int_{C_1}\frac{4zdz}{iz^2(4+(z+\frac{1}{z})^2)}=\int_{C_1}\frac{4zdz}{i(4z^2+(z^2+1)^2)}=\int_{C_1}\frac{4zdz}{i((2z-(z^2+1)i)(2z+(z^2+1)i))}=\int_{C_1}\frac{4zdz}{(z^2+2iz+1)(iz^2+2z+i)}$$
For the contour $C_1$, a circle of radius $1$ about the origin, traversed once. When I find the roots of the polynomials on the bottom, I get: $-i\pm i\sqrt{2}$ and $i \pm i\sqrt{2}$.
Then, I find that $$Res(-i+i\sqrt{2})=\frac{1}{2\sqrt{2}}$$
$$Res(i-i\sqrt{2})=\frac{1}{2\sqrt{2}}$$
$$\implies \int_{0}^{2\pi}\frac{1}{1+\cos^2(\theta)}d\theta=2\pi i \times (\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}})=\pi i \sqrt{2}$$
However, shouldn't the value of this integral be real since the original integral is real-valued?
| The same approach but with only one residue to evaluate:
\begin{align*}
\int_{0}^{2\pi}\frac{d\theta}{1+\cos^2(\theta)}&=\int_{0}^{2\pi}\frac{2d\theta}{3+\cos(2\theta)}=\int_{0}^{4\pi}\frac{dt}{3+\cos(t)}=2\int_{0}^{2\pi}\frac{dt}{3+\cos(t)}\\&=\frac{4}{i}\int_{|z|=1}\frac{dz}{6z+z^2+1}
=8\pi\,\mbox{Res}\left(\frac{1}{6z+z^2+1},-3+2\sqrt{2}\right)\\&=
\frac{8\pi}{6+2(-3+2\sqrt{2})}=\pi\sqrt{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.